Class $\mathbf{M}$.

In the class, there are 15 boys and 15 girls. On March 8, some boys called some girls to congratulate them on the holiday (no boy called the same girl twice). It turned out that the children could be uniquely divided into 15 pairs so that in each pair there was a boy who called the girl. What is the maximum number of calls that could have been made?
Let's denote the boys as $M_{1}, M_{2}, \ldots, M_{15}$, and the girls as $-D_{1}, D_{2}, \ldots, D_{15}$, such that $M_{1}-D_{1}, M_{2}-D_{2}, \ldots, M_{15}-D_{15}$ is the only pairing that satisfies the condition of the problem. Suppose each boy called at least two girls. We will draw an arrow from each girl $D_{i}$ to the boy $M_{i}$ she is paired with, and from each boy $M_{i}$ to another (different from $D_{i}$) girl he called. Then, from each child, there is one arrow. If we follow the arrows (starting from any girl), we will eventually reach a girl who has already appeared in the chain being constructed. Thus, there is a cycle in the corresponding graph. We will combine in this cycle each boy with the girl to whom he has an arrow; the other pairs will remain unchanged. We have obtained another partition into pairs, which contradicts the condition.

Therefore, there will be a boy who called exactly one girl. If we remove this pair, the number of calls will decrease by no more than 15 - the maximum possible number of calls this girl could have received. After this, there will again be a boy who made exactly one call to one of the remaining girls. Removing this pair, we will reduce the number of calls by no more than 14, and so on. In total, no more than $15+14+\ldots+2+1=120$ calls were made.

Exactly 120 calls are made, for example, if each girl $D_{i}$ received calls from boys $M_{1}, M_{2}, \ldots, M_{i}$.

## Answer

120 calls.