1.27 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 $k$.
(5th American Invitational Mathematics Examination, 1987)
[Solution] Transform the given inequality to $\frac{15}{8}>\frac{n+k}{n}>\frac{13}{7}$.
That is,
$$
\begin{aligned}
\frac{6}{7} & <\frac{k}{n}<\frac{7}{8} \\
48 n & <56 k<49 n .
\end{aligned}
$$

Thus, the problem is converted to finding the largest open interval $(48 n, 49 n)$, such that this interval contains only one multiple of 56.

Since the length of the interval $(48 n, 49 n)$ is $n$, it contains $n-1$ integers. If $n-1 \geqslant 2 \cdot 56$, then the interval must contain at least two multiples of 56, so we have
$$
\begin{array}{c}
n-1<2 \cdot 56 \\
n \leqslant 112 .
\end{array}
$$

When $n=112$, we have
$$
48 \cdot 112=56 \cdot 96<56 \cdot 97<56 \cdot 98=49 \cdot 112 .
$$

Here, $k=97$ is the only integer that satisfies the given inequality.
Therefore, $n=112$ is the solution.