ABSURD THINGS
White's in an impossible check from Qb6 and Rg4.
One of white's Rooks is promoted, since Ra1 died in a cage. The promoting pawn could only have been the f2 or g2 pawn. Combine that with the doubled h-pawns, and white requires 3 P captures to get this position.
Black's only missing Bc8 and Bf8, both of which were killed at home. There are no pieces available to sacrifice to the white pawns.
Black's two B's are both promoted. The c7 pawn needs 3 captures to promote on a white square. The h7 pawn needs 2. White's missing Q, N, Bf1, Ra1. This is one capture short of what black needs.
FIX:
Remove wNe4. Last move was 0...Rd4-g4+. There are now 5 missing white pieces, which is enough for black to promote the 2 pawns.
Remove bPg7. Now white only requires 2 captures, and Bf8 could have been captured on h4. The g7 pawn was taken by Pf2, allowing white to promote a pawn.
I'll leave c) - e) to other solvers. :)
Crazy situation, have anyone can solve this?
ABSURD THINGS
White's in an impossible check from Qb6 and Rg4.
One of white's Rooks is promoted, since Ra1 died in a cage. The promoting pawn could only have been the f2 or g2 pawn. Combine that with the doubled h-pawns, and white requires 3 P captures to get this position.
Black's only missing Bc8 and Bf8, both of which were killed at home. There are no pieces available to sacrifice to the white pawns.
Black's two B's are both promoted. The c7 pawn needs 3 captures to promote on a white square. The h7 pawn needs 2. White's missing Q, N, Bf1, Ra1. This is one capture short of what black needs.
FIX:
Remove wNe4. Last move was 0...Rd4-g4+. There are now 5 missing white pieces, which is enough for black to promote the 2 pawns.
Remove bPg7. Now white only requires 2 captures, and Bf8 could have been captured on h4. The g7 pawn was taken by Pf2, allowing white to promote a pawn.
I'll leave c) - e) to other solvers. :)
Continue if you can. I will post the final answer in next 2 weeks. Thank you :)
A) White - has a double check which is not possible only way would be to have the rook on g4 be on d4 than swing across;however, there is a knight in the way making it impossible to have a double check
White - has a rook that should not be there becuase if all pawns are counted from captures and on the board there is 8 accounted for so 1 rook is invalid
White - has a king in the captures which is impossible to have a king in capture becuase it would be in of the game.
Black - has a king in the captures which is impossible to have a king in capture becuase it would be in of the game.
Black - has 3 pieces that have been promoted knight, bishop, rook which are invalid becuase 2 of them exceed the allowed pawn count and 1 is impossible to promote because of the white pawns formation
B-E) This position is impossible to solve with only removing 2 pieces, because each side will have to remove more than a total of 2 pieces
White - remove 1 king from capture 1 rook
Black - remove 1 king from capture 1 rook(the g4 rook precisely) 1 bishop 1 knight than add 1 pawn to captures or on the board
In total you will have 6 pieces removed and 1 piece added than the position can be legal
sorry but each part has its own possible answer. Plz try again, thank you.
BigDoggProblem got (a) and (b) right. (Of course he did.) Solution is unique too.
(c): Focus on the impossible double check first. To resolve that, there are only 3 possibilities: remove -wNe4, -bQb6 or -bRg4. In each case, white must still have promoted one pawn to a rook (the wRa1 died in the cage still). Black in the diagram position is missing 2 bishops (useless; died on c8 and f8), 2 pawns (both promoted to light bishops) and nothing else. So white only captured the bishops on c8 and f8. Even removing bQb6 or bRg4 only accounts for white's doubled h-pawns - white's f-pawn doesn't have anything to capture and promote to a rook. Since in all 3 of these cases the position remains impossible, it is impossible to remove a single unit to yield a legal position.
(d) Again, to deal with the double check, one of the removed pieces must be the wNe4. We're not allowed to remove black pieces, so we know for a fact both black light bishops are promoted from the c7 and h7 pawns, and that white did not capture any black pieces but the c8- and f8-bishops (since there are the other black units on the board). So we must deal with the doubled h-pawns; remove one of the white h-pawns (say -wPh2). But we still have the fact that one of the white rooks is promoted (since wRa1 was captured in its cage) and white again cannot have captured anything with the f- or g-pawns to promote them. Therefore a contradiction, and removing any 2 white units will not result in a legal position.
I'll leave (e) up for grabs as that seems very combinatoric based, and I hate combinatorics. Although at a glance it's obvious that if the answer is that there are 10 ways, there must be only 5 other black pieces that can be removed in conjunction with [bQb6 or bRg4].
BigDoggProblem got (a) and (b) right. (Of course he did.) Solution is unique too.
(c): Focus on the impossible double check first. To resolve that, there are only 3 possibilities: remove -wNe4, -bQb6 or -bRg4. In each case, white must still have promoted one pawn to a rook (the wRa1 died in the cage still). Black in the diagram position is missing 2 bishops (useless; died on c8 and f8), 2 pawns (both promoted to light bishops) and nothing else. So white only captured the bishops on c8 and f8. Even removing bQb6 or bRg4 only accounts for white's doubled h-pawns - white's f-pawn doesn't have anything to capture and promote to a rook. Since in all 3 of these cases the position remains impossible, it is impossible to remove a single unit to yield a legal position.
(d) Again, to deal with the double check, one of the removed pieces must be the wNe4. We're not allowed to remove black pieces, so we know for a fact both black light bishops are promoted from the c7 and h7 pawns, and that white did not capture any black pieces but the c8- and f8-bishops (since there are the other black units on the board). So we must deal with the doubled h-pawns; remove one of the white h-pawns (say -wPh2). But we still have the fact that one of the white rooks is promoted (since wRa1 was captured in its cage) and white again cannot have captured anything with the f- or g-pawns to promote them. Therefore a contradiction, and removing any 2 white units will not result in a legal position.
I'll leave (e) up for grabs as that seems very combinatoric based, and I hate combinatorics. Although at a glance it's obvious that if the answer is that there are 10 ways, there must be only 5 other black pieces that can be removed in conjunction with [bQb6 or bRg4].
thanks you, I will show the final answer after 2 weeks :).
I'll leave c) - e) to other solvers. :)
Do you agree with Remellion 
I'll leave c) - e) to other solvers. :)
Do you agree with Remellion
Looks like solid reasoning to me.
The pieces above and below the board are not the captured pieces. They're from whatever board editor is being used in the screenshot. Ignore them and focus only on the position.
lol I don't get this at all. So I'm just going to wait for the answer peeps are paying attention to the checks but have not even mentioned the fact both sides have kings in there captures or the fact it is impossible for promotes to happen when the only open files are 2-3 diangoal spaces away from the missing pawns
My understanding of answer e) is that there are 10 ways to remove 2 pieces to make it all become a legal position? which if you remove both kings from either sides that is 2 pieces removed and it still not legal unless your saying 2 pieces from either side but even that makes no sense becuase there is more problems than that
It was a tiny misunderstanding, now you can do it again.
The maximum number of solutions for part e is 10, or it can be 9,8,7,6...
The pieces above and below the board are not the captured pieces. They're from whatever board editor is being used in the screenshot. Ignore them and focus only on the position.
thanks for your explaining 
Looks like solid reasoning to me.
Could you continue with part e, bro :)
Because of Whites pawn structure, the Bishop on c1 cannot have been moved yet and because of that and the pawns restricting him the White rook that started on a1 cannot possibly be out there.
No matter which of the two white rooks is the one originating from a1 he cannot possibly have reached the position he is in now legally, so he needs to be removed.
Looks like solid reasoning to me.
Could you continue with part e, bro :)
I probably could; I just don't want to. :)
Because of Whites pawn structure, the Bishop on c1 cannot have been moved yet and because of that and the pawns restricting him the White rook that started on a1 cannot possibly be out there.
No matter which of the two white rooks is the one originating from a1 he cannot possibly have reached the position he is in now legally, so he needs to be removed.
Don't forget the possibility of promotion to R.
I can take a shot at e), found 8 ways to remove two black pieces to fix the position. Let me know if there are mistakes in the solution.
Brief summary of the illegalities (more complete explanation in BiggDoggs post #2):
-White is in an impossible double check from the Qb6 and Rg4.
-3 black pieces have been captured by white pawns, 0 such captures available.
-5 white pieces are captured by black pawns, only 4 available.
Fix by removing two black pieces:
1. The double check must be fixed so one of the pieces removed must be Qb6 or Rg4.
2. Removal of any two-piece combination of the form Qb6/Rg4+X, where X is not a pawn, won't work. This because such a fix gives white only two pawn captures but does not change the pawn structure.
3. From 1 and 2 => the fix must be of the form "remove Qb6/Rg4+bP"
4. Combinations that do not work are:
Qb6/Rg4+Pa7: This fixes the double check, but not black pawn captures. Both black bishops are still promoted and require 5 captured white pieces.
Qb6/Rg4+Pe7: Leaves the black king in check.
5. Working combinations (one scenario given in each case):
Qb6/Rg4+Pb7: Pb7 captures white rook at a2 and promotes to N at a1. Pc7 captures 3 pieces and promotes to LSB, the other bB is the original LSB. 4 black pawn captures. The promoted N, h-pawn and Qb6/Rg4 give the required pawn captures for white.
Qb6/Rg4+Pd7: Pc7 requires 3 captures to promote to LSB, the other is again the original. The black h-pawn captures once and promotes to something at g1. Total black pawn captures 4. White f-pawn captures the promoted black h-pawn and the d-pawn and promotes to rook at d8. This leaves the Qb6/Rg4 to be captured at h4. Total 3 white pawn captures.
Qb6/Rg4+Pf7: Pf7 captures 2 white pieces and promotes to LSB. Ph7 does the same giving a total of 4 captured white pieces. Pf2 walks to f8 capturing nothing and promotes to rook. The white g-pawn captures Qb6/Rg4 at h4. Total 1 captured black piece.
Qb6/Rg4+Pg7: Pc7 captures 3 white pieces and promotes to LSB. Pg7 captures 1 white piece and promotes to LSB. Total 4 black pawn captures. Pf2 captures the black DSB, which is now free and promotes to rook at g8. Pg2 captures the h-pawn at h4. Total 2 white pawn captures.
6. Total number of "remove 2 black pieces" -fixes: 8.
BigDoggProblem got (a) and (b) right. (Of course he did.) Solution is unique too.
(c): Focus on the impossible double check first. To resolve that, there are only 3 possibilities: remove -wNe4, -bQb6 or -bRg4. In each case, white must still have promoted one pawn to a rook (the wRa1 died in the cage still). Black in the diagram position is missing 2 bishops (useless; died on c8 and f8), 2 pawns (both promoted to light bishops) and nothing else. So white only captured the bishops on c8 and f8. Even removing bQb6 or bRg4 only accounts for white's doubled h-pawns - white's f-pawn doesn't have anything to capture and promote to a rook. Since in all 3 of these cases the position remains impossible, it is impossible to remove a single unit to yield a legal position.
Could you explain more about -wNe4, I still dont understand why it's impossible to remove wNe4
Looks like solid reasoning to me.
Could you continue with part e, bro :)
I probably could; I just don't want to. :)
what do you think about Uncia_Uncia's solution :v
Now is the turn of White
a) Show all absurd things that happened in this (an absurd thing is: if two people playing a normal game of chess in accordance with current law, it can not occur like that).
b) Remove 2 pieces including 1 White and 1 Black to overcome all the absurd things in part a. Explain why?
c) Prove that if only one piece is removed, it can not overcome all the absurd things in part a.
d) Prove that if only 2 White pieces is removed, can not overcome all the absurd things in part a.
e) Prove that there are not more than 10 ways that: remove only 2 Black pieces to fix all the absurd things in part a.