Example 2 (2002 Shanghai Mathematics Competition) Given that the sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ are both arithmetic sequences, $S_{n}=a_{1}+a_{2}+\cdots+a_{n}, T_{n}=b_{1}+b_{2}+\cdots+b_{n}$, and for all positive integers $n, \frac{S_{n}}{T_{n}}=\frac{3 n+31}{31 n+3}$.
(1) Find the value of $\frac{b_{28}}{a_{28}}$;
(2) Find all positive integers $n$ such that $\frac{b_{n}}{a_{n}}$ is an integer.
Solution:
(1) Since $\frac{a_{n}}{b_{n}}=\frac{\frac{a_{1}+a_{2 n-1}}{2}}{\frac{b_{1}+b_{2 n-1}}{2}}=\frac{S_{2 n-1}}{T_{2 n-1}}=\frac{3(2 n-1)+31}{31(2 n-1)+3}=\frac{3 n+14}{31 n-14}$.

Therefore, $\frac{b_{28}}{a_{28}}=\frac{31 \times 28-14}{3 \times 28+14}=\frac{61}{7}$.
(2) Since $(3 n+14,3)=1$, to make $\frac{b_{n}}{a_{n}}=\frac{31 n-14}{3 n+14}$ an integer, it is necessary that $\frac{3(31 n-14)}{3 n+14}=31-$ $\frac{476}{3 n+14}$ is an integer. Thus, $(3 n+14) \mid 476$.
Since $476=2^{2} \times 7 \times 17,3 n+14 \geqslant 17$.
Therefore, $3 n+14=17,28,34,68,119,238,476$.
Since $n$ is an integer, $n=1,18,35,154$.
Comment: This is a type of problem suitable for the college entrance examination. While most students might solve it using a "brute force" method, students with a solid foundation in mathematics should be able to provide a clever solution using properties of sequences.