4. As shown in Figure 4, given $\angle A O M=60^{\circ}$, there is a point $B$ on ray $O M$ such that the lengths of $A B$ and $O B$ are both integers, thus $B$ is called an "olympic point". If $O A=8$, then the number of olympic points $B$ in Figure 4 is $\qquad$
4. 4 .

As shown in Figure 7, draw $A H \perp O M$ at point $H$.
Since $\angle A O M$
$=60^{\circ}$, and $O A$
$=8$, therefore,
$O H=4$,
$A H=4 \sqrt{3}$.
Let $A B=$
$m, H B=n$
($m, n$ are positive integers). Clearly, in the right triangle $\triangle A H B$, we have
$$
m^{2}-n^{2}=(4 \sqrt{3})^{2} \text {, }
$$

which simplifies to $(m+n)(m-n)=48$.
Since $m+n$ and $m-n$ have the same parity, the factorization of 48 can only be the following three:
$$
48=24 \times 2=12 \times 4=8 \times 6 .
$$

Thus, when $(m, n)=(13,11),(8,4),(7,1)$, there are Olympic points $B_{1}, B_{2}, B_{3}$.

Additionally, it is easy to see that the point $B_{3}$ symmetric to the line $A H$ is also an Olympic point, denoted as $B_{3}^{\prime}$.
Therefore, there are four Olympic points: $B_{1}, B_{2}, B_{3}, B_{3}^{\prime}$.