The second-largest divisor of $n$ is of the form $\frac{n}{p}$ where $p$ is the smallest prime that divides $n$.
The given condition gives $2013 = n + \frac{n}{p} = \frac{n}{p}(p+1)$. Therefore, $p+1$ is a divisor of $2013$ and thus odd. So, $p$ is $2$, the only even prime.
The equation now becomes $2013 = \frac{n}{2} \cdot 3$ which gives the unique solution $n = 1342$. $\square$