Let $n$ be an even positive integer and $W\_n$ be the class of all $n\times n$ matrices with entries from the set $\{-1,0,1\}$ such that all row sums and column sums are equal to $0$.


For each $M\in W\_n$, let $k(M)$ be the number of ones in $M$. I think the number of $M\in W\_n$ with $k(M)$ even is much bigger than the number of $M\in W\_n$ with $k(M)$ odd. Am I right?



The class of $(-1,0,1)$-matrices with all row sums and column sums equal to $0$
Yes, this should be true. Here is an explanation why. Equivalently, we want to assign weights $-1,0,1$ to the edges of $K\_{n,n}$ such that the sum of the weights at each vertex is $0$. One way to do this is to choose edge-disjoint cycles $C\_1, \dots, C\_\ell$ in $K\_{n,n}$ with $\bigcup\_{i=1}^\ell E(C\_i)=E(K\_{n,n})$. For each cycle $C\_i$ we either set all its edges to have weight $0$, or we alternate its edges between $1$ and $-1$. Note that there are two ways to make the edges alternating. Conversely, every assignment of weights $-1,0,1$ to the edges of $K\_{n,n}$ such that the sum of the weights at each vertex is $0$ can be decomposed in this way. 


Let $C\_1, \dots, C\_\ell$ be a cycle-partition of $E(K\_{n,n})$. Call $C\_i$ *half-even* if $|C\_i|/2$ is even and $C\_i$ *half-odd* if $|C\_i|/2$ is odd. Since $n$ is even, say $n=2k$, $K\_{n,n}$ has $4k^2$ edges. Working mod $4$, we conclude that there are an even number of half-odd cycles. As noted, there are $3$ ways to assign weights to each cycle $C\_i$. No matter how we assign weights to the half-even cycles, the total number of ones will always be even. Thus, the parity of the number of ones only depends on how we assign weights to the half-odd cycles $O\_1, \dots, O\_{2m}$. The number of ways to assign weights to $O\_1, \dots, O\_{2m}$ so that the total number of ones is even is 
$$
\binom{2m}{0}2^0+\binom{2m}{2}2^2+\dots + \binom{2m}{2m}2^{2m}.
$$ 
By the Binomial Theorem, this is equal to $\frac{3^{2m}+1}{2}$ which is slightly more than half of $3^{2m}$ (the total number of ways to assign weights to $O\_{1}, \dots, O\_{2m}$). 


Now, the reason there should be far more weightings with an even number of ones than an odd number of ones is because there are many cycle decompositions of $K\_{n,n}$ with no half-odd cycles at all. The weightings that are generated by these cycle decompositions will *always* have an even number of ones, while the other ones (by the argument above) still generate more weightings with an even number of ones than an odd number of ones. Note that it is possible for different cycle decompositions to generate the same weighting, so this does not constitute a proof. 


