We perform the linear transformation $(x, y) \rightarrow(x-y, x+y)$, which has the reverse transformation $(a, b) \rightarrow\left(\frac{a+b}{2}, \frac{b-a}{2}\right)$. Then the equivalent problem has a parabola has a vertical axis of symmetry, goes through $A=(0,200)$, a point $B=(u, v)$ in $S^{\prime}=\{(x, y) \mid x+y>0, x>y, y<200, x, y \in \mathbb{Z}, \text { and } x \equiv y \bmod 2\}$ and a new vertex $W=(w, 0)$ on $y=0$ with $w$ even. Then $\left(1-\frac{u}{w}\right)^{2}=\frac{v}{200}$. The only way the RHS can be the square of a rational number is if $\frac{u}{w}=\frac{v^{\prime}}{10}$ where $v=2\left(10-v^{\prime}\right)^{2}$. Since $v$ is even, we can find conditions so that $u, w$ are both even: $v^{\prime} \in\{1,3,7,9\} \Longrightarrow\left(2 v^{\prime}\right)|u, 20| w$, $v^{\prime} \in\{2,4,6,8\} \Longrightarrow v^{\prime}|u, 10| w$, $v^{\prime}=5 \Longrightarrow 2|u, 4| w$. It follows that any parabola that goes through $v^{\prime} \in\{3,7,9\}$ has a point with $v^{\prime}=1$, and any parabola that goes through $v^{\prime} \in\{4,6,8\}$ has a point with $v^{\prime}=2$. We then count the following parabolas: - The number of parabolas going through $(2 k, 162)$, where $k$ is a nonzero integer with $|2 k|<162$. - The number of parabolas going through $(2 k, 128)$ not already counted, where $k$ is a nonzero integer with $|2 k|<128$. (Note that this passes through $(k, 162)$.) - The number of parabolas going through $(2 k, 50)$ not already counted, where $k$ is a nonzero integer with $|2 k|<50$. (Note that this passes through $\left(\frac{2 k}{5}, 162\right)$, and any overlap must have been counted in the first case.) The number of solutions is then $2\left(80+\frac{1}{2} \cdot 64+\frac{4}{5} \cdot 25\right)=264$.