6. The largest positive integer $n$ for which the inequality $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds for a unique integer $k$ is
6.112

Analysis: The original problem is equivalent to the existence of a unique integer $k$ satisfying $\frac{6 n}{7} < k \leq \frac{7 n}{6}$. When $s > 96$, $\frac{7}{6} s \geq \frac{8 s+16}{7}$, hence there is no solution;
When $96 \geq s \geq 48$, $\frac{8 s+8}{7} \leq \frac{7}{6} s \leq \frac{8 s+16}{7} \leq \frac{7 s+7}{6}$, hence the solution is $\frac{7}{6} s \leq n \leq \frac{8 s+16}{7}$; when $47 \geq s \geq -1$, $\frac{7}{6} s \leq \frac{8 s+8}{7} \leq \frac{7 s+7}{6} \leq \frac{8 s+16}{7}$, hence the solution is $\frac{8 s+8}{7} \leq n \leq \frac{7 s+7}{6}$; when $-1 > s$, $\frac{8 s+8}{7} > \frac{7 s+7}{6}$, hence there is no solution; therefore, $n \leq 112$, when $s=96$, $n$ reaches its maximum value, at this time $k=97$.