Example 4 As shown in Figure 1, quadrilateral $ABCD$ is a rectangle. Two people, A and B, start from points $A$ and $B$ respectively at the same time, and move counterclockwise along the rectangle. In which circle does B possibly catch up with A for the first time? In which circle does B certainly catch up with A at the latest? Please explain your reasoning.
Explanation: Let $A D=B C=a \mathrm{~m}, A B=C D=b . \mathrm{m}$, and suppose the first time Yi catches up with Jia is after
$$
\frac{2 a+b}{74-65}=\frac{2 a+b}{9}
$$

minutes. The distance Yi has run when he first catches up with Jia is
$$
\frac{2 a+b}{9} \times 74(\mathrm{~m}) \text {. }
$$

At this point, the number of laps Yi has run is $p$, then
$$
\begin{array}{l}
p=\frac{74(2 a+b)}{9} \div(2 a+2 b) \\
=\frac{37(2 a+b)}{9(a+b)}=4+\frac{38 a+b}{9(a+b)} \\
=9-\frac{7 a+44 b}{9(a+b)} .
\end{array}
$$

Thus, when $4\frac{35}{2} b$, we have $\frac{7 a+44 b}{9(a+b)}<1$, so Yi will certainly catch up with Jia by the time he completes the 9th lap at most.

Note: The key to solving this problem is the skillful use of partial fraction decomposition.