You are required to find position such that:
1) Any move made by White or Black leads to stalemate for both sides immediately.
2) The position can be reached from the starting position of a chess game.
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We are looking for a diagram which can be part of a legal position with either White to move or Black to move. The further constraint is that any legal move by either side must result in diagram which is stalemate, whoever is on move. I am trying to maximize the number of pieces.
At least 30 pieces is possible:
[title "Move by either player would be immediate mutual stalemate"] [fen "brnbKRRN/qnk1pBN1/rb1pPpPp/p1pP1P1P/PpP5/8/1P6/8 w - - 0 1"]
Black g pawn captured the two missing (White) pieces to promote to dark-squared bishop, and also unblocked the g-file for the White pawn to advance. The only legal moves (whoever has the move) would be b3=, but the positions are dead, as would any which satisfy this challenge.
One tricky point is that if Black is on move, then White has many possible last moves, so by the en passant convention, Black is not permitted to capture bxa3ep or bxc3ep.
EDIT: As few as 8 pieces is possible.
[title "Move by either player would be immediate mutual stalemate"] [fen "5k1K/p3pP1P/4P3/P7/8/8/8/8 w - - 0 1"]
If Black moved last it must have been Ke8-f8, Ke8xNf8 or Ke8xBf8. White has many more choices for last move. But in either case the position is legal.
If you want a solution that doesn't rely on the en passant convention, here is a solution with 29 men that is heavily based on Laska's answer. The missing White knights ensure that one of the black pawns could have promoted.
When each sides moves their pawn to f3, the only legal move, a mutual stalemate shall occur.
[FEN "bqnbKRQB/rnk1pBRP/rb1pP1P1/p1pP4/PpP2p2/1P6/5P2/8 w - - 0 1"]
Just as a fun fact, the White dark-squared Bishop could easily be replaced by a knight and the position would still be a mutual stalemate after either side moves.
Here’s an alternative 8-piece solution.
[FEN "KBk5/P1P1p3/2P5/4P3/8/8/8/8 w - - 0 1"]
This shows that 2 pieces is the least that one side can have.