Hrvatska 2011

Let $p$ and $q$ be positive integers such that $\frac{2011}{2010}=\frac{p+1}{p}\cdot \frac{q+1}{q}$. Then $2011pq=2010(pq+p+q+1)$, i.e. $pq=2010(p+q+1)$.

From the last equation we find $$p=\frac{2010(q+1)}{q-2010}=\frac{2010(q-2010)+2010\cdot 2011}{q-2010}=2010+\frac{2010\cdot 2011}{q-2010}.$$ Since $p$ and $q$ are positive integers, it follows that $q-2010$ is a positive divisor of $2010\cdot 2011$. Each divisor of $2010\cdot 2011$ corresponds to exactly one pair $(p,q)$. Since $2010\cdot 2011=2\cdot 3\cdot 5\cdot 67\cdot 2011$, the number of its divisors is $2^5=32$. Finally, since the pairs $(p,q)$ and $(q,p)$ determine the same representation, the number of required representations is 16.