II. (16 points)
Given that the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ 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.
$$
\begin{array}{l}
\text { II. (1) } \frac{a_{n}}{b_{n}}=\frac{\frac{a_{1}+a_{2 n-1}}{2}}{\frac{b_{1}+b_{2 n-1}}{2}}=\frac{S_{2 n-1}}{T_{2 n-1}}=\frac{3(2 n-1)+31}{31(2 n-1)+3} \\
=\frac{3 n+14}{31 n-14}, \\
\end{array}
$$

Therefore, $\frac{b_{28}}{a_{28}}=\frac{31 \times 28-14}{3 \times 28+14}=\frac{31 \times 2-1}{3 \times 2+1}=\frac{61}{7}$.
(2) $\because(3 n+14,3)=1, \therefore \frac{b_{n}}{a_{n}}=\frac{31 n-14}{3 n+14}$ is an integer $\Leftrightarrow \frac{3(31 n-14)}{3 n+14}=31-\frac{476}{3 n+14}$ is an integer
$$
\begin{array}{l}
\Leftrightarrow(3 n+14) \mid 476 . \\
\because 476=2^{2} \times 7 \times 17,3 n+14 \geqslant 17, \\
\therefore 3 n+14=17,28,34,68,119,238,476 .
\end{array}
$$
$\because n$ is an integer, $\therefore n=1,18,35,154$.