13. Given that $n$ and $k$ are natural numbers, and satisfy the inequality $\frac{9}{17}<\frac{n}{n+k}<\frac{8}{15}$. If for a given natural number $n$, there is only one natural number $k$ that makes the inequality true, find the maximum and minimum values of the natural number $n$ that meet the condition.
When $n=10, 11, \cdots, 16$, respectively, we have
$$
\begin{array}{l}
8 \frac{3}{4}<k<8 \frac{8}{9}, 9 \frac{5}{8}<k<9 \frac{7}{9}, \\
10 \frac{1}{2}<k<10 \frac{2}{3}, 11 \frac{3}{8}<k<11 \frac{5}{9}, \\
12 \frac{1}{4}<k<12 \frac{4}{9}, 13 \frac{1}{8}<k<13 \frac{1}{3}, \\
14<k<14 \frac{2}{9},
\end{array}
$$

None of these intervals contain an integer $k$.
When $n=17$, we get $14 \frac{7}{8}<k<15 \frac{1}{9}$. Thus, $k=15$.
It is evident that the maximum value of the natural number $n$ that meets the condition is 144, and the minimum value is 17.