8. Find the largest positive integer $n$, such that the inequality $\frac{8}{15}<$ $-\frac{n}{n+k}<\frac{7}{13}$ holds for exactly one integer $k$.
8 . Transform the original inequality to $\frac{15}{8}>\frac{n+k}{n}>\frac{13}{7}$, which is equivalent to $\frac{7}{8}>\frac{k}{n}>\frac{6}{7} \Leftrightarrow 49 n>56 k>48 n$. Therefore, the problem is converted to finding the largest open interval ( $48 n$, $49 n$ ) that contains only one multiple of 56. Since the interval length is $n$, it contains $n-1$ integers. If $n-1 \geq 2 \times 56$, then the interval must contain at least 2 multiples of 56 $\Rightarrow n-1<2 \times 56 \Rightarrow n \leqslant 112$. When $n=112$, we have $48 \times 112=56 \times 96<56 \times 97<56 \times 98=49 \times 112$. Here, $k=97$. It is clearly the only integer that satisfies this inequality, so $n=112$ is the solution.