Take any rational number $r$ that is close enough to $\sqrt{2001}$. Then for all primes $p$ we easily find nonnegative $m_p,n_p$ with $2m_p-3n_p=v_p(r)$ (here again $v_p$ the $p$-adic valuation). For all but finitely many $p$ (namely those that occure in $r$) we can set $m_p=n_p=0$. Then we set $m=\prod_{p \in \mathbb P} p^{m_p}$ and $n=\prod_{p \in \mathbb P} p^{n_p}$, done.

Peter wrote: Prove that there exist positive integers $ m$ and $ n$ such that \[ \left\vert\frac{m^{2}}{n^{3}}-\sqrt{2001}\right\vert <\frac{1}{10^{8}}.\] Reference for this problem 'The Grosman Meomorial Mathematical Olympiad 1999' is wrong. The correct reference is (V. Bernik) Belarus 2001.