Answer: 280.

Solution: Estimation. Suppose fewer than 280 games have passed, i.e., more than 20 games remain. Then, among the participants of the "White Rook" school, there are at least two students who have not yet played all their games. In this case, Sasha cannot accurately name the participant of the next game from the "White Rook" school. Similarly, Sasha cannot name the participant of the next game from the "Black Bishop" school. Therefore, $n$ must be at least 280.

Example. We have just proven that if $n<280$, Sasha cannot name either of the participants of the next game. But is there a situation when $n=280$ where Sasha can still name someone who will play in the next game? It turns out, yes! For this, the 14 students of the "White Rook" school must play all their games, which is exactly 280 games. Then, in the next game, the remaining student from the "White Rook" school will definitely participate - and this is the one Sasha will name.

## 9th Grade
8-8. A chess tournament is held according to the following system: each of the 15 students from the "White Rook" school must play one game with each of the 20 students from the "Black Bishop" school, i.e., a total of 300 games should be played. At any given time, no more than one game is played.

After $n$ games, a spectator named Sasha, who has watched all the games and knows all the participants, exclaims: "I can definitely name one of the participants of the next game!" What is the smallest $n$ for which this could happen?