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{array}{rl}\sum _{i=1}^{n}i{x}_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{array}$$ \quad \square