AustriaMO2013

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 $n=1342$. $\square$