8. Given a positive integer $n$. Then the positive integers $m$ for which there exist positive integers $x_{1}<x_{2}<\cdots<x_{n}$ satisfying
$$
\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=m
$$

are $\qquad$
8. $1,2, \cdots n$.

First, we have $m \leqslant \frac{1}{1}+\frac{2}{2}+\cdots+\frac{n}{n}=n$.
Take $x_{1}=n, x_{2}=2 n, \cdots, x_{n}=n^{2}$. Then $\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=1$.
Take $x_{1}=1, \cdots, x_{k}=k, x_{k+1}=(n-k)(k+1)$, $\cdots, x_{n}=n(n-k)$. Then
$$
\begin{array}{l}
\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=k+\frac{n-k}{n-k} \\
=k+1(k=1,2, \cdots, n-1) .
\end{array}
$$

Therefore, $m \in\{1,2, \cdots, n\}$.