How many ways are there to fill up a $2n \times 2n$ matrix with $1, -1$? How many ways are there to fill up a $2n \times 2n$ matrix with $1, -1$ so that each column and each row has exactly $n $ $1$'s and $n$ $-1$'s  ?
I tried for cases $n=1 , 2$ but the solutions were just case checking so I can't apply it to the general case.
 A: I see there's an algorithmic way of doing this by computer to arrive at http://oeis.org/A058527.
This is not an answer but it is a suggestion of an approach how to solve algebraically and perhaps a partial answer that can be developed:
If we ignore temporarily the requirement that columns add up to zero, each row can satisfy the rules in this many ways: $$\binom{2n}{n}$$
So the number of ways the entire matrix can fulfil the "rows sum to zero" rule is $$\binom{2n}{n}^{2n}$$
Now if we can multiply this by the probability that every column obeys the "columns sum to zero" rule, given that the rows rule is obeyed, we will have the answer.
Since every row contains exactly half 1's and -1's, the probability the first column obeys the columns rule is given by the binomial probability of exactly $n$ successes in $2n$ trials:
$$P_{R1}=\binom{2n}{n}\cdot 2^{-2n}$$
The probability the 2nd column sums to zero (given that column 1 is now fixed) is again the probability of $n$ successes in $2n$ trials, except now half of our rows have a $-1$ eliminated and half have a $+1$ eliminated, so this is given by:
$$P_{R2}=\binom{2n}{n}\left(\frac{n}{2n-1}\right)^n\left(\frac{n-1}{2n-1}\right)^n$$
So if we can continue this process through to $P_{R2n}$ the answer to your question will be:
$$\binom{2n}{n}^{2n}\cdot\prod_{k=1}^{2n}P_{Rk}$$
Where $$P_{Rk}=\binom{2n}{n}\left(\frac{n}{2n-k}\right)^n\left(\frac{n-k}{2n-k}\right)^n$$
The bit I'm not totally clear on in my mind just now, is whether we can continue to generate $P_{R3...}$ through to $P_{R2n}$.  Clearly only $n$ terms are non-unitary, and not $2n$ as we need, because beyond that $k$ would exceed $n$ but if we're lucky this will be because once we have randomly selected the first $n$ columns, the remaining $n$ are fixed.  I think the symmetry of the problem gives us a reasonably good chance of this (by which I mean greater than 1 in 20!)
This can be checked by calculating the above function against known values for $P$ and if it matches, we know it's worthwhile investing in thinking about the probability of the $3rd$ and greater columns summing to zero.
If we're not lucky we would have to make a more in-depth model of the probabilities of columns fitting the "sum to zero" rule after the 2nd column, which in particular involves accounting for the fact that the 2nd $1$ chosen from any given row may have been taken from a row from which a $1$ was previously taken, or a row from which a $-1$ was previously taken, and so on for subsequent rows.  It's fairly obvious to see that e.g. in a 4-column table once we take two 1's from a row, that row is guaranteed to yield only $-1$'s in all subsequent columns.
