Observe that $n^6+31n^4-900=(n^2-5)(n^2+6)(n^2+30)$, and the facts $2009=7^2\cdot 41$, $2010=2\cdot 3\cdot 5\cdot 607$, and $2011$ is a prime. Hence, $2009\cdot 2010\cdot 2011 = 2\cdot 3\cdot 5\cdot 7^2\cdot 41\cdot 67\cdot 2011$. Take $n\equiv 0\pmod{30}$, which handles divisibility with $2\cdot 3\cdot 5$. Now for $7^2$, let $n\equiv 22\pmod{49}$. Then, $n^2+6=490$ is divisible by $49$ (to find this, note that neither $5$ nor $-30$ are quadratic residues modulo $7$, thus $49\mid n^2+6$, see $n\equiv 1\pmod{7}$, let $n=7k+1$ and work out the details). Finally, it remains to show, for each of the primes from the set $S=\{41,67,2011\}$, either $5$ or $-6$ or $-30$ are quadratic residues. If at least one is, we are done. Suppose the assumption is false, and, for instance $(\frac{5}{p})=(\frac{-6}{p})=(\frac{-30}{p})=-1$ for a $p\in S$. But note that $(\frac{-30}{p})=(\frac{5}{p})(\frac{-6}{p})$, which is a contradiction. With this, let $n_1,n_2,n_3$ be such that $n_1^2\in\{5,-6,-30\}\pmod{41}$, $n_2^2\in\{5,-6,-30\}\pmod{67}$, and $n_3^2\in\{5,-6,-30\}\pmod{2011}$. It now suffices to choose $n\equiv 0\pmod{30}$, $n\equiv 22\pmod{49}$, $n\equiv n_1\pmod{41}$, $n\equiv n_2\pmod{67}$, and $n\equiv n_3\pmod{2011}$. By the Chinese remainder theorem, such an $n$ indeed exists. Remark. Not only is there such an $n$, but there are infinitely many such $n$'s.

grupyorum wrote: Hence, $2009\cdot 2010\cdot 2011 = 2\cdot 3\cdot 5\cdot 7^2\cdot 41\cdot 607$. Typo?