Example 3. In the figure, the large circle is a 400-meter track, and the track from $A$ to $B$ is 200 meters long. The straight-line distance is 50 meters. A father and son start running counterclockwise from point $A$ along the track for a long-distance run. The son runs the large circle, while the father runs straight every time he reaches point $B$. The father runs 100 meters in 20 seconds, and the son runs 100 meters in 19 seconds. If they run at these speeds, how many times will they meet in 35 minutes from the start?
Let the number of laps the father and son run when they meet be $N_{1}, N_{2}\left(N_{1}, N_{2}\right.$ are positive integers), and let the distance from point $A$ to the meeting point on the left half-circle be $x$ meters $(0 \leqslant x \leqslant 200)$. According to the problem, we have
$$
\frac{250 N_{1}+x}{100 / 20}=\frac{400 N_{2}+x}{100 / 19} .
$$

This simplifies to $\frac{x}{200}=38 N_{2}-25 N_{1}$ being an integer. Since $0 \leqslant \frac{x}{200} \leqslant 1$, it can only be that $x=0$ or $x=200$. Therefore, their meeting points are at $A$ or $B$.
$\therefore 38 N_{2}-25 N_{1}-1$ or $38 N_{2}-25 N_{1}=1$.
Solving $38 N_{2}-25 N_{1}=0$ gives $N_{1}=38 k, N_{2}=$ $25 k .(k \in \mathbb{Z})$

Solving $38 N_{2}-25 N_{1}=1$ gives $N_{1}=38 k_{1}+3$, $N_{2}=25 k_{1}+2 .(k \in \mathbb{Z})$
Given that the father and son each run one lap in 50 and 76 seconds, respectively.

And $\left[\frac{2100}{50}\right)=42$ (laps), $\left[\frac{2100}{76}\right]=27$ (laps).
Their meeting at point $A$ can be determined by the solution to $38 N_{1}-25 N_{2}=0$.

Therefore, for $N_{1}=38 k, N_{2}=25 k$, we can take $k=$ 0,1 .

Similarly, for $N_{1}=38 k_{1}+3, N_{2}=25 k_{1}+2$, we can take $k_{1}=0,1$.
Thus, excluding the starting point, they meet 3 times. Answer: From the start of the race, they meet 3 times in 35 minutes.