1. (16 points) If $n$ is a positive integer greater than 2, find the minimum value of
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}
$$
When $n=3$,
$$
\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60} \text {. }
$$

Assume that when $n=k(k \geqslant 3)$,
$$
\sum_{i=k+1}^{2 k} \frac{1}{i} \geqslant \frac{37}{60} \text {. }
$$

Then when $n=k+1$,
$$
\begin{array}{l}
\sum_{i=n+1}^{2 n} \frac{1}{i}=\sum_{i=k+2}^{2 k+2} \frac{1}{i} \\
=\sum_{i=k+1}^{2 k} \frac{1}{i}+\frac{1}{2 k+1}+\frac{1}{2 k+2}-\frac{1}{k+1} \\
=\sum_{i=k+1}^{2 k} \frac{1}{i}+\frac{1}{2 k+1}-\frac{1}{2 k+2} \\
>\sum_{i=k+1}^{2 k} \frac{1}{i} \geqslant \frac{37}{60} .
\end{array}
$$

Therefore, the minimum value sought is $\frac{37}{60}$.