20. Pauli and Bor are playing the following game. There is a pile of 99! molecules. In one move, a player is allowed to take no more than 1% of the remaining molecules. The player who cannot make a move loses. They take turns, with Pauli starting. Who among them can win, regardless of how the opponent plays?
Answer: Pauli. Solution. Consider the game with the same rules for a pile of 99! - 1 molecules (let's call this the smaller game). Since the game will end in a finite number of moves and there are no ties, one of the players has a winning strategy. There are two possible cases.

(1) In the smaller game, the second player wins. Then in the original game, Pauli takes one molecule on his first move. This will reduce the situation to the smaller game, where he will move second, and therefore win.

(2) In the smaller game, the first player wins by taking $x$ stones on his first move. Then by the condition, $x \leqslant \frac{99!-1}{100}$. Since 99! is divisible by 100, $x+1 \leqslant \frac{99!}{100}$. This means that in the original game, Pauli has the right to take $x+1$ stones on his first move. Then he will end up in the situation after the winning move of the first player in the smaller game. Acting further as this player, he will win.