Problem 930

Answer: 120.

Solution. Note that $N$ is the total number of games that remain to be played in the tournament.

Let's describe an example where $N=120$. Number the students and schoolchildren from 1 to 15. Suppose the schoolchild with number $k$ needs to play with students numbered from 1 to $k$. Then the total number of games remaining to be played is

$$1+2+3+\ldots+15=120$$

games. It is not hard to verify that there is exactly one way to schedule 15 games in one day (the first schoolchild must play with the first student, the second with the second, the third with the third, ..., the fifteenth with the fifteenth).

Now we will prove that $N \leqslant 120$. Without loss of generality, we will assume that the only way to play 15 games is when the first schoolchild plays with the first student, the second schoolchild with the second student, ..., the fifteenth schoolchild with the fifteenth student. These 15 pairs will be called direct, and pairs of players with different numbers will be called cross.

Note that we cannot have a situation where the $k$-th schoolchild needs to play with the $m$-th student, and the $m$-th schoolchild needs to play with the $k$-th student (otherwise, there is another way to play 15 games). Thus, for each pair of numbers $k$ and $m$, no more than one cross game is scheduled. The total number of cross games is then no more than $\frac{15 \cdot 14}{2}=105$. Adding the direct games, we get no more than 120.