Example 5 Find the smallest positive integer $n$ such that for $n$, there exists a positive integer $k$ satisfying $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$.
From the given, we have $\frac{15}{8}>\frac{n+k}{n}>\frac{13}{7}$, which means $\frac{15}{8}>1+\frac{k}{n}>\frac{13}{7}$, so $\frac{6}{7}<\frac{k}{n}<\frac{7}{8}$. Taking $n=9$, we get $\frac{54}{7}<k<\frac{63}{8}$, with no integer value for $k$. Trying $n=10, n=11, n=12$, $n=13, n=14$, we get $\frac{60}{7}<k<\frac{70}{8}, \frac{66}{7}<k<\frac{77}{8}, \frac{72}{7}<k<\frac{84}{8}, \frac{78}{7}<k<\frac{91}{8}, \frac{84}{7}<k<\frac{98}{8}$, respectively, and $k$ cannot take an integer value in any of these cases. When $n=15$, $\frac{90}{7}<k<\frac{105}{8}$, $k$ can take 13, which satisfies the condition. Therefore, the minimum value of $n$ is 15.