If $n$ and $m$ are both odd, Vasyl wins. Take the point $Q$ with $x_Q = \frac{m+1}{2}$ and $y_Q=\frac{n+1}{2}$. Vasyl will mark first $Q$. Then, if Petryk marks a point $P$, Vasyl will mark its reflection with respect to $Q$. If $n$ is odd, $m$ is even, Petryk wins. Take the line $d$ with $x_d=\frac{m+1}{2}$ (the line perpendicular to $Ox$ in point ($\frac{m+1}{2}$, $0$)). If Vasyl marks a point $P$, Petryk will mark its reflection with respect to $d$. If $n$ is even, $m$ is odd, Petryk wins. Take the line $d$ with $y_d=\frac{n+1}{2}$ (the line perpendicular to $Oy$ in point ($0$, $\frac{n+1}{2}$)). If Vasyl marks a point $P$, Petryk will mark its reflection with respect to $d$. If $n$ and $m$ are both even, Petryk wins again. Take again the line $d$ with $x_d=\frac{m+1}{2}$ (the line perpendicular to $Ox$ in point ($\frac{m+1}{2}$, $0$)). If Vasyl marks a point $P$, Petryk will mark its reflection with respect to $d$. So, Vasyl wins iff $n$ and $m$ are both odd. If one of $n$ or $m$ is even, Petryk wins.