Given the relation:
\[\frac{1}{x}+\frac{1}{y} = \frac{1}{2010}, \: \: \: x,y \in \mathbb{N}.\: \: \: \: \: \: \: \: (1)\]Both $x$ and $y$ must be greater than $2010$.Let $x = 2010 + k$, for some $k \in \mathbb{N}$.Then $y$ can be expressed as: \[y = \frac{2010(2010+k)}{k} = 2010 + \frac{2010^2}{k}\]The question is, how many different natural numbers, $k$, divide the square of $2010$ (including the trivial case $k = 1$)?The prime factorization of the square of $2010$ is: $2010^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 67^2$. Thus, the
number of divisors, i.e. the number of $(x,y)$-pairs is: $3^4 = 81$.
The answer implies, that a specific pair, e.g. $(2011,2010\cdot 2011)$ and its permutation $(2010\cdot 2011, 2011)$ both count. Otherwise, the answer would be $41$ pairs.