Let $A\in\mathbb{R}^{n\times n}$ be an invertible 0-1 matrix. Is it possible that the sum $a:=\sum_{i,j}(A^{-1})_{ij}$ of entries of $A^{-1}$ is not equal to 1, but exponentially close (w.r.t. $n$) to 1?

More precisely, the *conjecture* is that if $a\geq 1$ then either $a=1$ or $|a|\geq 1+p(n)/q(n)$, for $p$ and $q$ fixed polynomials (some other not too fast growing functions might do, too).

**Remark.** The condition $a\geq 1$ is not redundant, as there exist matrices $A$ with $a<1$, e.g. $$A=\left(\begin{array}{rrrr}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1
\end{array}\right), \quad A^{-1}=\left(\begin{array}{rrrr}
-1 & -1 & 0 & 1 \\
-1 & -1 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{array}\right), \quad a=\sum_{i,j} (A^{-1})_{ij}=0.
$$

An (admittedly very shaky) evidence for this is that it is true if the row sums $r_i$ of $A$ are all equal to each other. Indeed, let $\overline 1$ denote the all-1 vector. Then $a=\sum_{i=1}^n x_i$ for $x$ satisfying $Ax=\overline 1$, and $$ \overline{1}^\top Ax=\sum_i r_i x_i=\overline{1}^\top \overline{1}=n,\quad \text{implying }\quad a\geq \frac{n}{n-1} $$