The answer is (C). First of all, note that a number has exactly $2011$ positive divisors if and only if it is of the form $p^{2010}$ with $p$ prime. We therefore need to solve $x^2 - y^2 = 2011^{2010}$.

By imposing that $x + y = p^{\alpha}$ and that $x - y = p^{\beta}$, with $\alpha + \beta = 2010$, we obtain that $x = \frac{p^{\alpha} + p^{\beta}}{2}$, that $y = \frac{p^{\alpha} - p^{\beta}}{2}$ and hence that $\alpha > \beta$ since $y$ must be positive. The equation $\alpha + \beta = 2010$ with $\alpha > \beta$ has exactly $1005$ solutions. Since $x$ and $y$ are uniquely determined by $\alpha$ and $\beta$, it follows that the solutions of the initial equation are also exactly $1005$.