SAMC

Let $n(n+2010)=k^2$ for some integer $k$.

We can write: $$n(n+2010)=k^2$$ $$n^2+2010n-k^2=0$$ This is a quadratic in $n$: $$n^2+2010n-k^2=0$$ The discriminant must be a perfect square for $n$ to be integer: $$\Delta =2010^2+4k^2=m^2$$ for some integer $m$.

So: $$m^2-4k^2=2010^2$$ $$(m-2k)(m+2k)=2010^2$$ Let $d$ be a positive divisor of $2010^2$, and set: $$m-2k=d,m+2k=\frac{2010^2}{d}$$ Then: $$m=\frac{d+\frac{2010^2}{d}}{2}=\frac{d+2010^2/d}{2}$$ $$2k=m+2k-m=\frac{2010^2}{d}-d$$ $$k=\frac{2010^2-d^2}{4d}$$ Now, $n$ is given by the quadratic formula: $$n=\frac{-2010\pm m}{2}$$ So: $$n=\frac{-2010\pm \frac{d+2010^2/d}{2}}{2}=\frac{-2010}{2}\pm \frac{d+2010^2/d}{4}$$ Thus, all integers $n$ for which $n(n+2010)$ is a perfect square are: $$n=\frac{-2010}{2}\pm \frac{d+2010^2/d}{4}$$ where $d$ is a positive divisor of $2010^2$ such that $d+2010^2/d$ is divisible by $4$ (so $n$ is integer).

Therefore, all such integers $n$ are given by the above formula for all positive divisors $d$ of $2010^2$ with $d+2010^2/d$ divisible by $4$.