Answer: 3.

Solution. Let's construct the following graph: vertices are players, edges are played matches. According to the condition, for both tournaments, this graph is the same. Consider the first tournament and select the matches where the winners had not won before (for example, the first match). Then the corresponding edges form a path, and the other edges are connected to this path by one end. In particular, if we remove all pendant vertices, only our path without the end vertices will remain.

Now consider the same graph as the graph of the cup tournament. If we remove the pendant vertices from it, a graph of the tournament on $2^{n-1}$ winners of the first stage will remain. It is obviously a path only for $n \leq 3$, otherwise, the tournament winner will have a degree of at least 3. Therefore, $n \leq 3$.

It remains to provide an example for $n=3$. Let the participants be numbered from 1 to 8 and the pairs in the cup be as follows (the loser is indicated first, the winner second): $1-2,3-4,5-6,7-8,2-4,6-8$, 4-8. Then in the knockout stage, the pairs could have been as follows (the winner is again indicated second): $1-2$, $2-4,3-4,4-8,7-8,8-6,6-5$.
1. On the first day, $2^{n}$ schoolchildren played ping-pong in a "knockout" format: first, two played, then the winner played with the third, the winner of this pair played with the fourth, and so on until the last schoolchild (there are no draws in ping-pong). On the second day, the same schoolchildren played for the cup: first, they arbitrarily split into pairs and played, the losers were eliminated, and the winners again arbitrarily split into pairs and played, and so on. It turned out that the sets of pairs that played on the first and second days were the same (possibly with different winners). Find the largest possible value of $n$.

B. Frenkin