China Southeastern Mathematical Olympiad

We see that if $2012 \le n \le 4023$, then $\sqrt{\frac{n-2011}{2012}} - \sqrt{\frac{n-2012}{2011}} \ge 0$ and $\sqrt[3]{\frac{n-2013}{2011}} - \sqrt[3]{\frac{n-2011}{2013}} < 0$.

Otherwise,
$$\begin{aligned} \sqrt{\frac{n-2011}{2012}} &\le \sqrt{\frac{n-2012}{2011}} &&\Leftrightarrow n > 4023 \\ \sqrt[3]{\frac{n-2013}{2011}} &\ge \sqrt[3]{\frac{n-2011}{2013}} &&\Leftrightarrow n \ge 4024. \end{aligned}$$
Thus, if $n \ge 4024$, then
$$\sqrt{\frac{n-2011}{2012}} - \sqrt{\frac{n-2012}{2011}} < 0 \le \sqrt[3]{\frac{n-2013}{2011}} - \sqrt[3]{\frac{n-2011}{2013}}$$
So the minimum of $n$ is $4024$. $\boxed{4024}$