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:
$$m - 2k = d, \quad m + 2k = \frac{2010^2}{d}$$
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 $4$.