8. If the inequality $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ is satisfied by only one integer $k$, then the maximum value of the positive integer $n$ is ( ).
(A) 100
(B) 112
(C) 120
(D) 150
8. B.

From the given inequality, we have
$$
\frac{13}{7}<\frac{n+k}{n}<\frac{15}{8} \text {, }
$$

then $\frac{6}{7}<\frac{k}{n}<\frac{7}{8}$,
so $\frac{6 n}{7}<k<\frac{7 n}{8}$.
Given that there is only one integer $k$ between $\frac{6 n}{7}$ and $\frac{7 n}{8}$, we have
$$
\frac{7 n}{8}-\frac{6 n}{7} \leqslant 2 \text {. }
$$

Solving this, we get $n \leqslant 112$.
When $n=112$, $96<k<98$, there exists a unique $k=97$.