9. (14 points) In the Cartesian coordinate system $x O y$, points with both integer coordinates are called integer points. Given $O(0,0), A(2,1)$, and $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$. If $S_{\triangle O M M}=3$, find the number of integer points $M$ that satisfy this condition.
9. Connect $O A$. It is easy to know that there are two integer points $M_{1}(-6,0)$ and $M_{2}(6,0)$ on the $x$-axis inside the ellipse that satisfy the problem.

Draw two lines $l_{1}$ and $l_{2}$ parallel to the line $O A$ through points $M_{1}$ and $M_{2}$, respectively.

According to the principle that triangles with equal bases and heights have equal areas, all integer points $M$ that meet the conditions lie on the lines $l_{1}$ and $l_{2}$. It is easy to know
$$
k_{O A}=\frac{1-0}{2-0}=\frac{1}{2} \text {. }
$$

Therefore, the equations of the lines $l_{1}$ and $l_{2}$ are
$$
y=\frac{1}{2}(x+6), y=\frac{1}{2}(x-6) \text {. }
$$

Given that $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$, we have
$$
\frac{x^{2}}{200}+\frac{y^{2}}{8}<1 \text {. }
$$

Solve $\left\{\begin{array}{l}\frac{x^{2}}{200}+\frac{y^{2}}{8}<1, \\ y=\frac{1}{2}(x+6)\end{array}\right.$ and $\left\{\begin{array}{l}\frac{x^{2}}{200}+\frac{y^{2}}{8}<1, \\ y=\frac{1}{2}(x-6),\end{array}\right.$ to get $-10<x<-\frac{10}{29}, \frac{10}{29}<x<10$.

Since $M$ is an integer point and lies on the lines $l_{1}$ and $l_{2}$, $x$ must be an even number.

Therefore, in $-10<x<-\frac{10}{29}$ and $\frac{10}{29}<x<10$, $x$ has 4 even numbers each.
Thus, the number of integer points that meet the conditions is 8.