Find all positive integers $n$ for which there are distinct integers $a_1$, ..., $a_n$ such that 

$\frac{1}{a_1}+\frac{2}{a_2}+...+\frac{n}{a_n}=\frac{a_1+a_2+...+a_n}{2}$.
To solve the problem, we need to find all positive integers \( n \) for which there exist distinct integers \( a_1, a_2, \ldots, a_n \) such that the equation 

\[
\frac{1}{a_1} + \frac{2}{a_2} + \cdots + \frac{n}{a_n} = \frac{a_1 + a_2 + \cdots + a_n}{2}
\]

holds true.

1. **Check for small values of \( n \):**

   - For \( n = 1 \):
     \[
     \frac{1}{a_1} = \frac{a_1}{2}
     \]
     Multiplying both sides by \( 2a_1 \):
     \[
     2 = a_1^2
     \]
     Since \( a_1 \) must be an integer, there is no solution for \( n = 1 \).

   - For \( n = 2 \):
     \[
     \frac{1}{a_1} + \frac{2}{a_2} = \frac{a_1 + a_2}{2}
     \]
     Multiplying both sides by \( 2a_1a_2 \):
     \[
     2a_2 + 4a_1 = a_1a_2(a_1 + a_2)
     \]
     This equation is difficult to satisfy with distinct integers \( a_1 \) and \( a_2 \), and no simple solutions exist.

2. **Check for \( n = 3 \):**

   - For \( n = 3 \):
     \[
     \frac{1}{a_1} + \frac{2}{a_2} + \frac{3}{a_3} = \frac{a_1 + a_2 + a_3}{2}
     \]
     One possible solution is \( a_1 = 1, a_2 = 2, a_3 = 3 \):
     \[
     \frac{1}{1} + \frac{2}{2} + \frac{3}{3} = 1 + 1 + 1 = 3
     \]
     \[
     \frac{1 + 2 + 3}{2} = \frac{6}{2} = 3
     \]
     Thus, \( n = 3 \) is a solution.

3. **Check for \( n = 4 \):**

   - For \( n = 4 \):
     \[
     \frac{1}{a_1} + \frac{2}{a_2} + \frac{3}{a_3} + \frac{4}{a_4} = \frac{a_1 + a_2 + a_3 + a_4}{2}
     \]
     One possible solution is \( a_1 = 1, a_2 = 2, a_3 = -3, a_4 = 4 \):
     \[
     \frac{1}{1} + \frac{2}{2} - \frac{3}{3} + \frac{4}{4} = 1 + 1 - 1 + 1 = 2
     \]
     \[
     \frac{1 + 2 - 3 + 4}{2} = \frac{4}{2} = 2
     \]
     Thus, \( n = 4 \) is a solution.

4. **Generalization for \( n \geq 3 \):**

   If a solution exists for a given \( n \), a solution exists for \( n + 4 \). This can be shown by choosing \( k \in \mathbb{N} \) such that \( k(n+1) > \max_{i \in [1,n]} |a_i| \). Then, the sequence \( (a_1, a_2, \ldots, a_n, k(n+1), -k(n+2), -k(n+3), k(n+4)) \) is a solution for \( n + 4 \).

   This implies that if a solution exists for \( n = 3 \) or \( n = 4 \), then solutions exist for all \( n \geq 3 \).

Hence, the result is:

\[
\boxed{n \in [3, +\infty)}
\]