I know for a fact that for any skew-symmetric matrix $A$, there cannot be a vector $x\geq 0$ so that $Ax\geq e$, where $e$ is the vector of all $1$'s. However, the proof I have is very indirect and is based on the connection of (2-person zero-sum) game theory with the primal-dual theory of linear programs.
Is there a more direct proof?
I assume that by "a vector $x \geq 0$" you mean a vector with entries in $\mathbb{R}_{\geq 0}$, and "$Ax \geq e$" means the entries of the vector $Ax$ are all real numbers greater than or equal to $1$. In particular, I assume that you are working over $\mathbb{R}$ throughout. Please let me know if I have misunderstood.
Assuming my understanding is correct, note $x \cdot Ax = A^Tx \cdot x = -A x \cdot x = -x \cdot Ax$, and so in particular $x \cdot Ax = 0$. On the other hand, if $x \geq 0$ and $Ax \geq e$, then $x \cdot Ax \geq x \cdot e \geq 0$. This vanishes only when $x=0$, which would force $Ax = 0 \not\geq e$.