On his page on Harry Goldsteen's "horse concoction", Tim Krabbé gives this position by "N. Höeg, 1935":
With the stipulation "add 10 white rooks; black mates in 1". He then claims that "Retrograde-analysis will prove that these Rooks can only stand on a1, a2, a7, b4, c1, c2, c5, c7, d7 and f7, allowing Black to mate with Qxb4." This is the claimed solution:
[FEN "4k3/RpRR1Rpp/p1pQ1p2/1qRpp3/1R6/8/RKR5/R1R5 b - - 0 1"] [StartFlipped "0"]
However, I don't see how such the supposed solution position could be legal. All of White's pawns have promoted, but because all of Black's pawns are present they must all have captured at least once to do so. But Black has only 5 missing pieces.