 Solution

Strictly positive entries imply $v>0$. If $q$ is an optimal strategy, set $x=q/v$. Then $Ax\le 1$ and $1^{\mathsf{T}}x=1/v$. Conversely, a nonzero feasible $x$ with $s=1^{\mathsf{T}}x$ gives a strategy $q=x/s$ satisfying ${\max }_i(Aq{)}_i\le 1/s$. Thus $v\le 1/s$, or $s\le 1/v$. The optimal objective is therefore $s_{\ast }=1/v$, andA nonzero feasible point exists by taking a sufficiently small positive multiple of a probability vector, and positivity of $A$ bounds the feasible objective, so the optimum is attained.

For general entries choose a constant $c$ so that $A^{\prime }=A+c11^{\mathsf{T}}$ is strictly positive. Every pair of strategies has its payoff increased by exactly $c$, so optimal strategies are unchanged. Solve the positive problem for $A^{\prime }$, recover $v^{\prime }=1/s_{\ast }^{\prime }$, and then $\boxed{v=v^{\prime }-c}$ for the original game.