Find all positive integers $n$ such that the set $S=\{1,2,3, \dots 2n\}$ can be divided into $2$ disjoint subsets $S_1$ and $S_2$, i.e. $S_1 \cap S_2 = \emptyset$ and $S_1 \cup S_2 = S$, such that each one of them has $n$ elements, and the sum of the elements of $S_1$ is divisible by the sum of the elements in $S_2$.

[i]Proposed by Viktor Simjanoski[/i]
We need to find all positive integers \( n \) such that the set \( S = \{1, 2, 3, \ldots, 2n\} \) can be divided into two disjoint subsets \( S_1 \) and \( S_2 \) with \( |S_1| = |S_2| = n \), and the sum of the elements in \( S_1 \) is divisible by the sum of the elements in \( S_2 \).

1. **Define the sums of the subsets:**
   Let \( \sum_{i \in S_1} i = A \) and \( \sum_{i \in S_2} i = B \). Since \( S_1 \cup S_2 = S \) and \( S_1 \cap S_2 = \emptyset \), we have:
   \[
   A + B = \sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2} = n(2n+1)
   \]
   We are given that \( A \mid B \).

2. **Establish bounds for \( A \) and \( B \):**
   Note that \( A \geq 1 + 2 + \ldots + n = \frac{n(n+1)}{2} \) and \( B \leq 2n + (2n-1) + \ldots + (n+1) = \frac{n(3n+1)}{2} \). Therefore:
   \[
   B \leq \frac{n(3n+1)}{2} < \frac{3n(n+1)}{2} \leq 3A
   \]
   Since \( A \mid B \), this implies that \( B \in \{A, 2A\} \).

3. **Case 1: \( B = A \):**
   If \( B = A \), then:
   \[
   A = \frac{n(2n+1)}{2}
   \]
   For \( A \) to be an integer, \( n(2n+1) \) must be even, which implies \( n \) must be even. For all even \( n \), we can construct \( S_1 \) as follows:
   \[
   S_1 = \{1, 2n\} \cup \{2, 2n-1\} \cup \ldots \cup \left(\frac{n}{2}, 2n+1-\frac{n}{2}\right)
   \]
   It is straightforward to check that \( |S_1| = n \) and \( A = n(2n+1) \).

4. **Case 2: \( B = 2A \):**
   If \( B = 2A \), then:
   \[
   A = \frac{n(2n+1)}{3}
   \]
   For \( A \) to be an integer, \( 3 \mid n(2n+1) \). Since \( 2n+1 \) is always odd, \( n \) must not be congruent to 2 modulo 3, i.e., \( n \not\equiv 2 \pmod{3} \).

5. **Intermediate sums:**
   Consider the collection \( \mathcal{F} \) of all sets \( X \subseteq \{1, 2, \ldots, 2n\} \) such that \( |X| = n \). The minimum sum of the elements of a set in \( \mathcal{F} \) is:
   \[
   m = 1 + 2 + \ldots + n = \frac{n(n+1)}{2}
   \]
   and the maximum sum is:
   \[
   M = 2n + (2n-1) + \ldots + (n+1) = \frac{n(3n+1)}{2}
   \]
   We claim that all intermediate sums in \([m, M]\) can be achieved by a set in \( \mathcal{F} \). Assume all sums in \([m, t]\) have been achieved for some \( t \geq m \). If \( t = M \), we are done. If not, we want to find a set that achieves \( t+1 \). Let \( T = \{x_1, \ldots, x_n\} \) be a set such that its elements sum to \( t \).

   If there are two elements of \( T \) that are not consecutive, we may increment the smallest one of them by one and finish. Moreover, if \( x_n \neq 2n \), we may increment \( x_n \) by one and finish. If neither of these happens, set \( T \) must necessarily be \( \{n+1, n+2, \ldots, 2n\} \), which is a contradiction as we assumed \( t \neq M \).

6. **Conclusion:**
   The working \( n \) are the evens and the \( n \not\equiv 2 \pmod{3} \), that is all \( n \not\equiv 5 \pmod{6} \).

The final answer is \( \boxed{ n \not\equiv 5 \pmod{6} } \).