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, $$\begin{array}{rlrl}\sqrt{\frac{n-2011}{2012}}& \le \sqrt{\frac{n-2012}{2011}}& & \iff n>4023\\ \sqrt[3]{\frac{n-2013}{2011}}& \ge \sqrt[3]{\frac{n-2011}{2013}}& & \iff n\ge 4024.\end{array}$$ 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 $4024$. $\boxed{4024}$