From the given equation, we have $xy - nx - ny = 0 \Rightarrow (x-n)(y-n) = n^2$. Then, besides $x = y = 2n$, for any $x-n$ equal to a proper divisor of $n^2$, we will get a positive integer solution $(x, y)$ satisfying the required condition. Therefore, $n^2$ should have exactly 2010 proper divisors that are less than $n$.

Suppose $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$, where $p_1, \dots, p_k$ are prime numbers different from each other. Then the number of proper divisors of $n^2$ less than $n$ is
$$\frac{(2\alpha_1 + 1) \cdots (2\alpha_k + 1) - 1}{2}$$
So $(2\alpha_1 + 1) \cdots (2\alpha_k + 1) = 4021$. Since 4021 is prime, we get $k = 1, 2\alpha_1 + 1 = 4021$, and $\alpha_1 = 2010$.

Therefore, $n = p^{2010}$, where $p$ is any prime number.