Find the smallest positive integer $n$ such that there exists a positive integer $k$ for which $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds.
Solution: For the inequality $\frac{8}{15}1$, i.e., $m>8$.
Therefore, $n>56$.
When $n=7 m+p\left(m \in \mathbf{N}_{+}, p \in \mathbf{N}\right.$, and $1 \leqslant p$ $\leqslant 6$), we have
$$
\begin{array}{l}
6 m+\frac{6}{7} p0$, i.e., $m>p$.

Since $1 \leqslant p \leqslant 6$, to find the smallest $m$, when $p=1$, $m_{\text {min }}=2$. Thus,
$$
n_{\text {min }}=7 m_{\min }+p=7 \times 2+1=15 .
$$

When $n=15$, we have $\frac{90}{7}<k<\frac{105}{8}$, i.e.,
$$
12 \frac{6}{7}<k<13 \frac{1}{8} \text {. }
$$

Therefore, $k=13$.
Hence, the smallest positive integer $n$ that satisfies the condition is $15$.