$$ \newcommand{\1}{\mathbf{1}} \newcommand{\im}{\text{Im }} \newcommand{\Span}[1]{\langle #1 \rangle} \newcommand{\orth}[1]{ #1^{\perp}} $$
Let $A$ be an $n \times n$ skew-symmetric matrix. Define $\1 := (1, \dots, 1)$ and assume that $A^{-1}(\1)$ is non-empty. Show that $$ A^{-1}(\1) \subseteq \1^{\perp} $$ Namely, if $Aq = \1$ then $q \cdot \1 = 0$
I checked this by hand in low dimension, but don't have much of an idea how to proceed generally. An attempt could be that $$ Aq = \1 \Rightarrow Aq \cdot \1 = -q \cdot A\1 = n $$ but not much from here.
The key step is probably that $\im{A} = \ker{A}^{\perp}$. This implies $$ \begin{split} \1 \in \im{A} &\iff \Span{\1} \subseteq \im{A} \\ & \iff \orth{\Span{\1}} \supseteq \ker{A} \\ & \iff \forall q \in \ker{A}, \, q \cdot \1 = 0 \end{split} $$
The solution of the affine system $Aq = \1$ is given by a particular solution $+ \ker{A}$, so $q \cdot \1$ is at least constant on $A^{-1}(\1)$.
To give some context: this problem arises when one looks for formal equilibria in Evolutionary Game Theory, namely points $q$ fulfilling $Aq \in \Span{\1}$ and $q \cdot \1 = 1$ for some matrix $A$. Those points can be used to construct Hamiltonian functions for a class of evolutionary dynamics.
$A\mathbf{q} = \mathbf{1} \Rightarrow A\mathbf{q} \cdot \mathbf{q}= \mathbf{1} \cdot \mathbf{q}$, but the LHS vanishes by antisymmetry.