Answer is $\boxed{4024}$

Let $s = n-2011$. we can rewrite the inequation as: $\sqrt{\frac{s}{2012}}-\sqrt{\frac{s-1}{2011}}<\sqrt[3]{\frac{s-2}{2011}}-\sqrt[3]{\frac{s}{2013}}$ for $s<2013$: we have $\sqrt{\frac{s}{2012}}-\sqrt{\frac{s-1}{2011}}\geq 0>\sqrt[3]{\frac{s-2}{2011}}-\sqrt[3]{\frac{s}{2013}}$ for $s = 2013$: the inequality is obviously true. Conclusion: the least natural number is $s = 2013$. so: $n=s+2011=4024$

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