Call good a number $p$ for which there exist positive integers $x_1 < x_2 < \dots < x_n$ such that $\frac{1}{x_1} + \frac{2}{x_2} + \dots + \frac{n}{x_n} = p$.
Since $x_1, x_2, \dots, x_n$ are integers and $x_1 < x_2 < \dots < x_n$, we have $x_k \ge k$, so $\frac{k}{x_k} \le 1$, for all $k = 1, 2, \dots, n$. Then $\frac{1}{x_1} + \frac{2}{x_2} + \dots + \frac{n}{x_n} \le n$, so every good number, if there exists any, is between 1 and $n$.

Next, we will show that any integer $p \in \{1, 2, \dots, n\}$ is good. Obviously, $n$ is good (for $x_k = k$) and 1 is also good (take $x_k = kn$). For $2 \le p \le n-1$, we write:
$$\sum_{k=1}^{n} \frac{k}{x_k} = \left( \frac{1}{x_1} + \frac{2}{x_2} + \dots + \frac{p-1}{x_{p-1}} \right) + \left( \frac{p}{x_p} + \dots + \frac{n}{x_n} \right),$$
so it is enough to choose $x_1, x_2, \dots, x_n$ such that the first sum is equal to $p-1$, and the second sum is equal to 1. We can do that by setting $x_k = k$, for $k = 1, 2, \dots, p-1$ and $x_k = k(n-p+1)$, for $p \le k \le n$. Notice that $x_1 < x_2 < \dots < x_n$ in all cases, so the good numbers are indeed 1, 2, $\dots,$ $n$.