For each positive integer $ n$ find all positive integers $ m$ for which there exist positive integers $ x_1<x_2<...<x_n$ with:

$ \frac{1}{x_1}\plus{}\frac{2}{x_2}\plus{}...\plus{}\frac{n}{x_n}\equal{}m.$
1. **Claim**: All positive integers \( m \leq \frac{n(n + 1)}{2} \) can be written in the form \( \frac{1}{x_1} + \frac{2}{x_2} + \cdots + \frac{n}{x_n} = m \) for some positive integers \( x_1 < x_2 < \cdots < x_n \).

2. **Proof of Upper Bound**: 
   - The maximum value of \( \frac{k}{x_k} \) for integers \( k \) and \( x_k \) occurs when \( x_k = k \). Thus,
     \[
     \frac{1}{x_1} + \frac{2}{x_2} + \cdots + \frac{n}{x_n} \leq 1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}.
     \]
   - Therefore, \( m \) cannot exceed \( \frac{n(n + 1)}{2} \).

3. **Base Case**: 
   - For \( n = 1 \), the maximum value is \( 1 \), which can be obtained with \( x_1 = 1 \). Hence, the base case is true.

4. **Inductive Step**:
   - Assume that for some \( k \), there exist \( x_1, x_2, \dots, x_k \) such that all \( m \leq \frac{k(k + 1)}{2} \) can be obtained.
   - We need to show that for \( n = k + 1 \), all \( m \leq \frac{(k+1)(k+2)}{2} \) can be obtained.

5. **Constructing \( m \) for \( n = k + 1 \)**:
   - By adding the term \( \frac{k + 1}{x_{k+1}} \), we can still obtain any value in \( m \).
   - If \( m \neq 1 \), we can write for some \( x_1, \dots, x_k \),
     \[
     \frac{1}{x_1} + \frac{2}{x_2} + \cdots + \frac{k}{x_k} = m - 1.
     \]
     Then,
     \[
     \frac{1}{x_1} + \frac{2}{x_2} + \cdots + \frac{k}{x_k} + \frac{k + 1}{x_{k+1}} = m - 1 + \frac{k + 1}{x_{k+1}}.
     \]
     Setting \( x_{k+1} = k + 1 \) gives the desired value for \( n \).

6. **Special Case for \( m = 1 \)**:
   - If \( m = 1 \), we can set \( x_1 = x_2 = \cdots = x_{k+1} = \frac{(k+1)(k+2)}{2} \).

7. **Range of Values**:
   - To obtain all values in the range \( \frac{k(k + 1)}{2} \) to \( \frac{k(k + 1)}{2} + (k + 1) \), we note that if we wish to obtain some value \( a \), we can find
     \[
     \frac{1}{x_1} + \frac{2}{x_2} + \cdots + \frac{k}{x_k} = a - (k + 1)
     \]
     so long as \( 1 \leq a - (k + 1) \leq \frac{k(k + 1)}{2} \).
   - Setting \( x_{k+1} = 1 \) gives
     \[
     \frac{1}{x_1} + \frac{2}{x_2} + \cdots + \frac{k+1}{1} = a - (k + 1) + \frac{k+1}{1} = a.
     \]
   - Thus, we can obtain all the values in that range.

8. **Conclusion**:
   - By induction, for any \( n \), we can obtain any \( m \leq \frac{n(n + 1)}{2} \).

\(\blacksquare\)

The final answer is \( \boxed{ m \leq \frac{n(n + 1)}{2} } \).