What are the positive integers $n$ such that $n+2009 \mid n^{2}+2009$ and $n+2010 \mid n^{2}+2010$?
On a $n+2009 \mid n^{2}+2009-(n+2009)=n^{2}-n$ and similarly $n+2010 \mid n^{2}-n$. Since $n+2009$ and $n+2010$ are consecutive and thus coprime, it follows that $(n+2009)(n+2010) \mid n^{2}-n$. Given $n \geqslant 0$, we have $(n+2009)(n+2010)>n^{2}-n$. The only possibility is therefore that $n^{2}-n=0$, which means $n=0$ or $n=1$, both of which are indeed solutions.