14+14. Wer gewinnt?
The pawn formation around bBh6 tells us that the black captures must
have been h7xQg6 and e7xBf6, accounting for all missing white men. White,
on the other hand, has made pawn captures bxa and gxh, the first taking
a pawn and the latter a rook. Both these pawns, as well as the original
h-pawn, have then promoted. Promotion to a bishop has taken place on
a8, while both h-pawns have been promoted to a knight, presently existing
on g1 and h3.
The task is then to unpromote wNg1 and wNh3 on h8, retract the resulting
pawns beyond h7 and then retract black's h7xQg6, after which the position
releases comparatively easily.
Suppose now that white has the move. Retroplay would be:
-1. ... Rf6-g6 -2. Nf4-h3 Rg6-f6 -3. Nh3-g1 Rf6-g6 -4. Ng6-f4
Nd6-b5 -5. Rb5-b7 Nb7-d6 -6. Nh8-g6 Rg6-h6 -7. h7-h8=N Rf6-g6 -8. Nh3-f4
Rg6-f6 -9. f2-f3 Rf6-g6 -10. Ng6-f4 Nb6-d5 -11. Rd5-d7 Nd7-b6 -12. Nh8-g6
g6-g5 and white is retro-stalemate.
So black has the move. Retroplay is then -1. Nf4-h3 Rf6-g6
-2. Ng6-f4 Nd6-b5 -3. Rb5-b7 Nb7-d6 -4. Nh8-g6 Rg6-f6 -5. h7-h8=N Rf6-g6
-6. Nh3-g1 Rg6-f6 -7. Nf4-h3 Rf6-g6 -8. Ng6-f4 Nb6-d5 -9. Rd5-d7 Nd7-b6
-10. Nh8-g6 g6-g5 -11. f2-f3 Bf4-h6 -12. h6-h7 Be3-f4 -13. h5-h6 Bf4-e3
-14. h7-h8=N Bd6-f4 -15. h6-h7 h7xQg6 etc, etc.
The solution must then be: Black to move and win with 1. ...
Re6 2. Ne6 Ne7#
Solution by mr.mip