Of course for odd $n$ this number is just 0, so suppose $n = 2m$ is even. For each $k$, the number of paths with $k$ steps up is precisely $\binom{2m}{k, k, m - k, m - k}$: we must choose $k$ of $2m$ steps to be up, $k$ to be down, $m - k$ to be right and $m - k$ to be left. Thus, the total number of such paths is \[ \sum_{k = 0}^m \binom{2m}{k, k, m - k, m - k} = \binom{2m}{m} \sum_{k = 0}^m \binom{m}{k}^2 = \binom{2m}{m}^2. \]

A more combinatorial way to obtain the binomial coefficient squared is to rotate the plane 45 degrees so that the steps are $(\pm \sqrt{2}/2, \pm \sqrt{2}/2)$ with the $\pm$s chosen independently. In $2m$ moves, we need $m$ plusses and $m$ minuses in the x-component, and likewise $m$ plusses and $m$ minuses in the y-component. There are $\binom{2m}{m}$ ways to choose these for each axis, getting a total of $\binom{2m}{m}^2$. This doesn't generalize as well beyond 2 dimensions though.