Alas, AI indicates that the answer is yes.
It is geometrically possible because even though there are only ten 'edible' pieces, enough captures can create two passed pawns, and that can be arranged.
It's definitely not normal though!
Here's how I saw it:
1 Why not? Both underpromoted 8 pawns
2 Maybe not. To promote, a pawn must bypass its opposing pawn; meaning captures
3. 20 knights! 16 Ps 4 Ns are involved
4 16 promotions required, but only 10 remaining pieces are 'edible'
5 The answer is probably no.
🤔