Given some invertible matrix $A$ with zero diagonal, and all-one vector $j$, when is $A^{-1}j \geq 0$ (elementwise)?
So far, I only found conditions for all entries of the inverse to be non-negative (in the context of $M$-matrices). But this is too strong for my purposes, and violates the zero-diagonal condition.
Of course, when $A$ has constant row-sum $k > 0$, the inverse row sums must be $1/k$. Is there some way to use this fact to say something about general matrices? If this is too hard to answer in general: Does assuming symmetry and/or non-negativity of $A$ help?
The following will help if you can adjust your problem to allow $A$ to have a positive diagonal: $A^{-1}$ has nonnegative row sums if $A$ is nonsingular and its transpose $A^T$ is a $B_0$-matrix (Lemma 4 in Christensen, 2019).
$A^T$ is a $B_0$-matrix if the column sums of $A$ are nonnegative and at least as large as any off-diagonal entry in the same column. In notation, let $A=(a_{ij})$ be an $n \times n$ matrix and define $c^+_j:=\max\{0,a_{ij}|i \neq j\}$ for $j=1,...,n$. Then $A^T$ is a $B_0$-matrix if, for $j=1,...,n$, $$\frac{1}{n} \sum_{i=1}^n a_{ij} \geq c^+_j. $$
Similarly, the inverse of $A$ has nonnegative column sums if $A$ is a $B_0$-matrix. A $B_0$-matrix must have nonnegative diagonal entries, however.
See also Pe$\tilde{n}$a, 2001.