Belarusian Mathematical Olympiad

Answer: $n = 39$.
(Solution of A. Semchankau, A. Zhuk.) Let
$$2013n = a^3 - b^3. \qquad (1)$$
Then
$$(1) \Leftrightarrow 61 \cdot 11 \cdot 3n = 2013n = (a-b)^3 + 3ab(a-b) \Rightarrow (a-b) \vdash 3,$$
i.e. $(a-b) = 3k, k \in \mathbb{N}$. So $61 \cdot 11 \cdot 3n = 3^3 k^3 + 3^2 abk$, whence $n \vdash 3$, i.e. $n = 3m, m \in \mathbb{N}$. Then (1) can be rewritten as
$$3k^3 + abk = 61 \cdot 11m. \qquad (2)$$
Note that if $k = 11, b = 10, a = 43$, then $m = 13$, i.e. $n = 39$. Show that $n = 39$ is the smallest possible value of $n$, i.e. $m = 13$ is the smallest possible value of $m$. Indeed, from (2) it follows that $k(3k^2 + 3kb + b^2) \vdash 11$, and it is easy to show that $k \vdash 11$. But for $k \ge 22$ we have $m > 13$.

Answer: $n=39$. (Solution of A. Semchankau, A. Zhuk.) Let
$$2013n = a^3 - b^3. \qquad (1)$$
Then
$$(1) \Leftrightarrow 61 \cdot 11 \cdot 3n = 2013n = (a-b)^3 + 3ab(a-b) \Rightarrow (a-b) \ge 3,$$
i.e. $(a-b) = 3k, k \in \mathbb{N}$. So $61 \cdot 11 \cdot 3n = 3^3 k^3 + 3^2 abk$, whence $n \ge 3$, i.e. $n = 3m, m \in \mathbb{N}$. Then (1) can be rewritten as
$$3k^3 + abk = 61 \cdot 11m. \qquad (2)$$
Note that if $k=11, b=10, a=43$, then $m=13$, i.e. $n=39$. Show that $n=39$ is the smallest possible value of $n$, i.e. $m=13$ is the smallest possible value of $m$. Indeed, from (2) it follows that $k(3k^2+3kb+b^2) \ge 11$, and it is easy to show that $k \ge 11$. But for $k \ge 22$ we have $m > 13$.