3. Connect the endpoints of arcs exceeding $120^{\circ}$. Let the number of line segments drawn from point $A$ be the maximum, denoted as $k$. Since in any three points on the circle, at least one arc does not exceed $120^{\circ}$, there are no triangles in this figure. Therefore, the $k$ endpoints of the $k$ line segments drawn from point $A$ are not connected to each other. Let the total number of line segments in the figure be $m$, then
$$
m \leqslant(21-k) k=-k^{2}+21 k=-\left(k-\frac{21}{2}\right)^{2}+\frac{441}{4} .
$$

When $k=10$, $m$ reaches its maximum value, which is 110.
Thus, there are at most 110 arcs exceeding $120^{\circ}$, and at most $C_{21}^{2}-110=100$ arcs not exceeding $120^{\circ}$.
On the other hand, divide the 21 points into two groups, with 10 and 11 points respectively, each group at one end of a diameter (the points within each group can be very close to each other), in this way, the number of arcs not exceeding $120^{\circ}$ is exactly $C_{10}^{2}+C_{11}^{2}=100$.
Three. (50 points)
There are 21 points on a circle. Among all the arcs with these 21 points as endpoints, what is the minimum number of arcs that do not exceed $120^{\circ}$?