, 2012

Note that
$$\sum_{n=0}^{1006} \frac{2012!}{(n!(1006-n)!)^2} = \frac{2012!}{(1006!)^2} \sum_{n=0}^{1006} \left( \frac{1006!}{n!(1006-n)!} \right)^2 = \left( \frac{2012!}{1006!} \right) \sum_{n=0}^{1006} \left( \frac{1006!}{n!} \right)^2.$$
Since $\binom{1006}{n} = \binom{1006}{1006-n}$, we have
$$\sum_{n=0}^{1006} \left(\binom{1006}{n}\right)^2 = \sum_{n=0}^{1006} \binom{1006}{n} \binom{1006}{1006-n} = \binom{2012}{1006},$$
by the Vandermonde identity. Therefore,
$$\sum_{n=0}^{1006} \frac{2012!}{(n!(1006-n)!)^2} = \left(\binom{2012}{1006}\right)^2$$
which is a perfect square.