Solved with AN1729 If $y\leq x$, then choose $w=142857$. Now notice that Malay will always move his pawn upwards and will not lose. So now we consider $x<y$. Claim - Only $(x,y) = (x,2x), (x,\frac{3x}{2}), (x,3x)$ work. Let the pawns on $(0,0)$ be $s_1$, $(x,0)$ be $s_2$ and $(y,w)$ be $m$. Let the vertical line on which $m$ moves be $k$. Let $h_1$ and $h_2$ be the directed vertical distances between $s_1$ and $s_2$, and $s_1$ and $m$ respectively. Let $l$ be the line formed by $s_1$ and $s_2$. Let $l \cap k$ be $W$. Proof - Assume $y \neq 2x, \frac{3x}{2}, 3x$, and that Sunaina can force a win. WLOG assume $(x,y) = 1$, and choose $w$ sufficiently large such that Malay does not lose on move $1$. Consider the penultimate position, where it is Malay to move. So, any move Malay makes should result in a loss in atmost one more move. Also in the ultimate position, $h_1$ must be a multiple of $x$ for $W$ to be a lattice point. Case 1 - $W$ is a lattice point in the penultimate position. Case 1a - $x \neq 1$. Then, Malay can always choose the point that is not $W$, and after Sunaina's move, $W$ will not be a lattice point, so Malay cannot lose. Case 1b - $x=1 \implies y>3$. So, Sunaina's next move will either move $W$ up or down by distance $y$ or up and down by distance $y-1$. Since $y>3$, Malay will always have a safe square and choose his move such that he will not get trapped. Case 2 - $W$ is not a lattice point in the penultimate position. ($x \neq 1$) Case 2a - Now, assume $x>2$. Then, Malay cannot lose on his move, and for $W$ to be a lattice point after Sunaina's move, $h_1 \equiv \pm 1 \pmod{x}$. Since $x>2$, Sunaina has to either increase or decrease $h_1$ (which depends on $h_1 \pmod{x}$). There is only one possible move for each of $s_1$ and $s_2$ to make $W$ a lattice point. Now, the $2$ positions that $W$ can reach on the next move are exactly $1$ unit apart. But, Malay can move his point such that after his move, $m$ will be on neither of these points, since Malay can choose between points having a gap of $2$. Case 2b - $x=2, y>3$. Now, since $W$ is not a lattice point, Sunaina can move both pawns up and down to make it a lattice point. So, $W$ can reach $4$ points. If the top point has coordinate $k$, then the other points have coordinates $k-1, k-y+1, k-y$. None of these have a distance of $2$ between them as $y>3$, so Malay will always have a safe square. Now assume $y = 2x, \frac{3x}2, 3x$. In each of these cases, $2$ of Sunaina's moves will give $W$ a distance of exactly $2$ between them, so Sunaina can trap Malay. It is trivial to see that Sunaina always reaches Malay (since the slope of $l$ changes by $\frac yx$ which is $>1$).

Solved with Nimloth149131215208 Denote $\frac{y}{x}$ as $r$. Note that $r>0$. I claim the answer is $r\in\{2,\frac{3}{2},3\}$.