China Girls' Mathematical Olympiad

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.