By replacing 2013 with 1, 2, 3, and 4 and then calculating the sum, you get a hunch for which number squared this sum yields. The factorials suggest that you should look in the binomial coefficients. We prove something more general:

$$\sum_{n=0}^{m} \frac{(2 m)!}{(n!(m-n)!)^{2}}=\binom{2 m}{m}^{2}$$

From this, the desired result follows.
It holds that

$$\frac{(2 m)!}{(n!(m-n)!)^{2}}=\frac{(m!)^{2}}{(n!(m-n)!)^{2}} \cdot \frac{(2 m)!}{(m!)^{2}}=\binom{m}{n}^{2} \cdot\binom{2 m}{m}$$

Thus, it is sufficient to prove that

$$\sum_{n=0}^{m}\binom{m}{n}^{2}=\binom{2 m}{m}$$

We do this combinatorially. Consider $2 m$ balls numbered from 1 to $2 m$, where the balls from 1 to $m$ are colored blue and the balls from $m+1$ to $2 m$ are colored red. You want to choose a total of $m$ balls. This can be done in $\binom{2 m}{m}$ ways. On the other hand, we can also first choose $n$ blue balls, with $0 \leq n \leq m$, and then choose $m-n$ red balls. This is equivalent to first choosing $n$ blue balls and then not choosing $n$ red balls. Thus, the number of ways to choose $m$ balls is also equal to

$$\sum_{n=0}^{m}\binom{m}{n}^{2}$$

Therefore, this sum is equal to $\binom{2 m}{m}$. And with that, we have proven (1).