There will be ${64 \choose n} \cdot {64-n \choose n}$ such combinations, since there are ${64 \choose n}$ ways to choose a subset of $n$ of the 64 variables to be -1, and then ${64-n \choose n}$ ways to choose a subset of $n$ of the $64-n$ remaining variables to be 1. Equivalently, this is ${64! \over n!^2 (64-2n)!}$.
You probably won't want to explicitly generate all of those combinations, as it would take exponentially long to do so (as there are exponentially many such combinations).
But if you really wanted to explicitly enumerate all combinations, this formula also highlights the way to do that. First, enumerate all possible ways to choose $n$ of the 64 variables; there are standard algorithms for how to do that (e.g., https://stackoverflow.com/q/127704/781723, https://en.wikipedia.org/wiki/Combination#Enumerating_k-combinations). Then, for each such, enumerate all possible ways to choose $n$ of the remaining $64-n$ variables (again, using the same algorithm). That will give you all ways to choose a combination of the sort you desire.
Still, any time you are enumerating exponentially many possibilities and then doing something to each, you should check whether maybe there's a better way to achieve your ultimate goal without enumerating all of them. If you want to find the combination that minimizes some cost function, there might be a better way to do that, depending on the cost function.
