Let the "depth" of a point $(a, b)$ be $(a+b)/2$. In our walk, the points must all have integer depth. The step $(1, 1)$ increases the depth by 1, and the steps $(1, -1)$ and $(-1, 1)$ both keep the depth constant. For any integer $k$ with $0 \leq k < n$, consider when we first move from depth $k$ to depth $k+1$. There are exactly $2k+1$ points possible as the source: they are the points $(0, 2k), (1, 2k-1), \ldots, (i, 2k-i), \ldots, (2k, 0)$. All of them yield a valid $(1, 1)$ step. Furthermore once we've chosen each of these points, the remaining keeping-depth-constant moves can be filled in in exactly one way. Therefore, the number of ways is $(2n-1)(2n-3)(2n-5)\ldots(1) = (2n-1)!!$