55rd Ukrainian National Mathematical Olympiad - Fourth Round

If integer number $n$ satisfies the given condition, then $4$ can be presented as product of four pairwise distinct integers. Since the integer divisors of this number are only $\pm 1$, $\pm 2$ and $\pm 4$, we have that sought-for divisors are $\pm 1$ and $\pm 2$. Indeed, if the absolute value of one of divisors is equal to $4$ then others are not less than $1$ by absolute value, a contradiction. Since $n-2013$ is the biggest factor, it should be equal to $2$. Also we see that $n=2015$ satisfies the condition of the problem.