Saudi Arabia Mathematical Competitions 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 $\left(\genfrac{}{}{0ex}{}{1006}{n}\right)=\left(\genfrac{}{}{0ex}{}{1006}{1006-n}\right)$, we have $$\sum _{n=0}^{1006}{\left(\left(\genfrac{}{}{0ex}{}{1006}{n}\right)\right)}^2=\sum _{n=0}^{1006}\left(\genfrac{}{}{0ex}{}{1006}{n}\right)\left(\genfrac{}{}{0ex}{}{1006}{1006-n}\right)=\left(\genfrac{}{}{0ex}{}{2012}{1006}\right),$$ by the Vandermonde identity. Therefore, $$\sum _{n=0}^{1006}\frac{2012!}{(n!(1006-n)!{)}^2}={\left(\left(\genfrac{}{}{0ex}{}{2012}{1006}\right)\right)}^2$$ which is a perfect square.