Problem 702

4. 197.

As shown in Figure 6, through point $P$, draw
$P I, P E, P F, P G$ perpendicular to sides
$O A, A B, B C, O C$ at $D, E$,
$F, G$. It is easy to see that
$$\begin{array}{l} P F+P D=100, \\ P E+P G=100 \text {. } \\ \text { If| } S_{\text {эхи }} \cdot S_{\text {эми: }} \\ =S_{\triangle P U} \cdot S_{\triangle M K C} \text {, } \\ \end{array}$$

we know $P D \cdot P F=P E \cdot P G$, which means
$$P D(100-P D)=P G(100-P G) \text {. }$$

Simplifying, we get $(P D-P G)(P D+P G-100)=0$.
Thus, $P D=P G$ or $P D+P G=100$, which means
$$P D=P G \text { or } P G=P F \text {. }$$

Therefore, $P$ is a point on the diagonal $O B$ or on the diagonal $A C$.

Similarly, when $P$ is a point on the diagonal $O B$ or on the diagonal $A C$, it satisfies
$$S_{\triangle m} \cdot S_{\triangle Y W:}=S_{\triangle M M} \cdot S_{\triangle J N:} \text {. }$$

Thus, $P$ is a good point if and only if $P$ is an internal lattice point on the diagonal $O B$ or the diagonal $A C$.

It is easy to see that there are 99 good points on the diagonal $O B$, and there are also 99 good points on the diagonal $A C$. It is also easy to see that the intersection point of the diagonals $O B$ and $A C$ is also a good point. Therefore, the number of good points that satisfy the condition is $99+99-1=197$.