Let $n$ be a positive integer. A finite set of integers is called $n$-divided if there are exactly $n$ ways to partition this set into two subsets with equal sums. For example, the set $\{1, 3, 4, 5, 6, 7\}$ is $2$-divided because the only ways to partition it into two subsets with equal sums is by dividing it into $\{1, 3, 4, 5\}$ and $\{6, 7\}$, or $\{1, 5, 7\}$
and $\{3, 4, 6\}$. Find all the integers $n > 0$ for which there exists a $n$-divided set.

Proposed by [i]Martin Rakovsky, France[/i]
To solve the problem, we need to find all positive integers \( n \) for which there exists an \( n \)-divided set. A set is \( n \)-divided if there are exactly \( n \) ways to partition it into two subsets with equal sums.

1. **Base Cases:**
   - For \( n = 0 \), the set \(\{1\}\) is \(0\)-divided because there are no ways to partition it into two subsets with equal sums.
   - For \( n = 1 \), the set \(\{1, 2, 3\}\) is \(1\)-divided because the only way to partition it into two subsets with equal sums is \(\{1, 2\}\) and \(\{3\}\).

2. **General Case for \( n \geq 2 \):**
   - Consider the set \( S = \{3n, 3n+1, \ldots, 4n-1, 4n+1, 4n+2, \ldots, 5n, 8n(n-2)\} \).
   - The sum of all elements in \( S \) is:
     \[
     \sum_{k=3n}^{4n-1} k + \sum_{k=4n+1}^{5n} k + 8n(n-2)
     \]
     We can calculate these sums separately:
     \[
     \sum_{k=3n}^{4n-1} k = \frac{(4n-1)(4n) - (3n-1)(3n)}{2} = \frac{16n^2 - 4n - 9n^2 + 3n}{2} = \frac{7n^2 - n}{2}
     \]
     \[
     \sum_{k=4n+1}^{5n} k = \frac{(5n)(5n+1) - (4n)(4n+1)}{2} = \frac{25n^2 + 5n - 16n^2 - 4n}{2} = \frac{9n^2 + n}{2}
     \]
     Adding these sums together:
     \[
     \frac{7n^2 - n}{2} + \frac{9n^2 + n}{2} + 8n(n-2) = 8n^2 - 16n
     \]
     Therefore, the total sum of the set \( S \) is \( 16n^2 - 16n \).

3. **Partitioning the Set:**
   - We need to partition \( S \) into two subsets with equal sums, each subset having a sum of \( 8n^2 - 8n \).
   - Place \( 8n(n-2) \) in one subset. The remaining elements in this subset must sum to \( 8n \).
   - The smallest element in \( S \) is \( 3n \), so there can be at most two other elements in this subset.
   - The only option is to choose pairs of the form \( \{4n - \ell, 4n + \ell\} \) for \( \ell = 1, 2, \ldots, n \).

4. **Conclusion:**
   - For each \( n \geq 2 \), the set \( S \) can be partitioned in exactly \( n \) ways as described.
   - Therefore, for \( n \geq 2 \), there exists an \( n \)-divided set.

The final answer is \( \boxed{ n \geq 2 } \).