Well, that's reaaally hard!!

$2013|a^3 - b^3 => 3.11.61|a^3 - b^3 => 3.11.61|(a - b)(a^2+ab+b^2)$

$3|a^3 - b^3 => 3|a-b$

$11|a^3 - b^3 => 11|a-b$

if $61|a-b => 2013|a-b$ and the smallest $n$ is $\frac{(2*2013)^3 - 2013^2}{2013}$ else $61|a^2+ab+b^2$ and $33|a-b$ so $b = a + 33k$ . for $k=1 61|(a+33)^2+a(a+33)+a^2 <=> 61|a^2 + 33a - 3 => a^2 + 33a - 3 = 61p <=>a^2 + 33a - 3-61p=0$ Suppose there are integer solutoins for $a =>$ discriminant is perfect square of odd integer. For 1, 3, 5 there is no sulution. For discriminant equal $7^2 => n = \frac{43^3 - 10^3}{2013}=39$