Saudi Arabian Mathematical Competitions

Let $a_i$ be the number of clubs that have $i$ members. Here, $2 \leq i \leq n$ and
$$m = a_2 + a_3 + \cdots + a_n.$$
We will count the tuples $(A, B, C)$ in which the students $A, B$ take part in the same club $C$ that has $i$ members.

1. The first way of counting:
- Choosing $1$ club among the clubs that have $i$ members, we have $a_i$ ways.
- Choosing $2$ students that take part in that club, we have $\binom{i}{2}$ ways.

2. The second way of counting:
- Choosing $2$ students among all students of the school, we have $\binom{n}{2}$ ways.
- Choose $1$ club in which these two students take part. By assumption, there is at most $1$ such club.

So we get the following relation:
$$a_i \binom{i}{2} \leq \binom{n}{2} \Leftrightarrow a_i \leq \frac{\binom{n}{2}}{\binom{i}{2}}.$$
Hence, we have
$$m = a_2 + a_3 + \cdots + a_n \leq \binom{n}{2} \left( \frac{1}{\binom{2}{2}} + \frac{1}{\binom{3}{2}} + \cdots + \frac{1}{\binom{n}{2}} \right).$$
Note that
$$\sum_{i=2}^{n} \frac{1}{\binom{i}{2}} = \sum_{i=2}^{n} \frac{2}{i(i-1)} = 2 \sum_{i=2}^{n} \left( \frac{1}{i-1} - \frac{1}{i} \right ) = 2 \left( 1 - \frac{1}{n} \right ) = \frac{2(n-1)}{n}.$$
Therefore,
$$m \leq \frac{n(n-1)}{2} \cdot \frac{2(n-1)}{n} = (n-1)^2.$$
$\square$