Berpow S.l.

At the alumni meeting, 45 people attended. It turned out that any two of them who had the same number of acquaintances among those present were not acquainted with each other. What is the maximum number of pairs of acquaintances that could have been among those who attended the meeting?
Since $45=1+2+3+\ldots+9$, we can divide 45 people into groups of $1,2, \ldots, 9$ people. Let people in the same group not know each other, while people in different groups know each other. Then each person in the $k$-th group has $45-k$ acquaintances.

Under this condition, the problem is solved, and the total number of pairs of acquaintances is

$C_{45}^{2}-\left(C_{2}^{2}+C_{3}^{2}+\ldots+C_{9}^{2}\right)=870$

We will prove that there could not have been a greater number of acquaintances. Fix some $k, 0 \leq k \leq 44$. Suppose some graduate knows exactly $k$ others. By the condition, none of his acquaintances can have exactly $k$ acquaintances.

Therefore, the number of graduates who know exactly $k$ others does not exceed

$45-k$.

Let $A_{0}, A_{1}, \ldots, A_{44}$ denote the number of graduates who have $0,1, \ldots, 44$ acquaintances, respectively. As shown above, $A_{k} \leq 45-k$, and moreover, $A_{0}+A_{1}+\ldots+A_{44}=45$.

We estimate the total number of acquaintances: $2 S=A_{1}+2 A_{2}+\ldots+44 A_{44}=A_{44}+\left(A_{44}+A_{43}\right)+\ldots+\left(A_{44}+A_{43}+\ldots+A_{36}\right)+\ldots+$

## Answer

870 pairs.