China Girls' Mathematical Olympiad

From the hypothesis, for any two boys, there is at most one girl that knows both of them. Let $x_i$ be the number of girls that know exactly $i$ boys, $1 \le i \le n$. So $\sum_{i=1}^{n} x_i = m$. By counting the number of the above two boys-one girl combinations, we have
$$\sum_{i \ge 2} \frac{i(i-1)}{2} x_i \le \frac{n(n-1)}{2}.$$
The number of boy-girl pairs that know each other is then
$$$$
$$\begin{aligned} \sum_{i=1}^{n} ix_i &= m + \sum_{i=2}^{n} (i-1)x_i \le m + \sum_{i=2}^{n} \frac{i(i-1)}{2}x_i \\ &\le m + \frac{n(n-1)}{2}. \end{aligned}$$ $\quad \square$