Romanian Mathematical Olympiad

$$S(n) = \{(k, d) \mid d \text{ divides } k, k \le d^2 \le n^2, 1 \le k \le n^2\}.$$
For $1 \le k \le n^2$, let $A(k)$ be the set of the natural divisors $d$ of $k$ such that $k \le d^2 \le n^2$.
A natural number $d \in \{1, 2, \dots, n\}$ belongs to the sets $A(d)$, $A(2d)$, $\dots$, $A(d^2)$ and only to them.
It follows that the contribution of each $d$ in the sum $\sum_{k=1}^{n^2} a(k) = \sum_{k=1}^{n^2} |A(k)|$ is $\underbrace{1 + 1 + \dots + 1}_{d \text{ terms}} = d$.
Thus, $\sum_{k=1}^{n^2} a(k) = \sum_{d=1}^{n} d = n(n+1)/2$.