There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \ldots , w_n$. It is known that if any coin with weight $ w_k$, where $ 1 \leq k \leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists.
1. **Base Case for \( n = 7 \):**
   - Consider the set \(\{1, 3, 5, 7, 9, 11, 13\}\).
   - We need to show that removing any coin allows the remaining coins to be split into two groups of equal weight.

   Let's verify this for each coin:
   - If we remove \(1\):
     \[
     13 + 11 = 3 + 5 + 7 + 9 \quad \text{and} \quad (5 + 9 + 11) - (3 + 7 + 13) = 2
     \]
   - If we remove \(3\):
     \[
     13 + 9 + 1 = 5 + 7 + 11 \quad \text{and} \quad (11 + 13) - (1 + 5 + 7 + 9) = 2
     \]
   - If we remove \(5\):
     \[
     13 + 9 = 1 + 3 + 7 + 11 \quad \text{and} \quad (3 + 7 + 13) - (1 + 9 + 11) = 2
     \]
   - If we remove \(7\):
     \[
     13 + 3 + 5 = 1 + 9 + 11 \quad \text{and} \quad (9 + 13) - (1 + 3 + 5 + 11) = 2
     \]
   - If we remove \(9\):
     \[
     13 + 7 = 1 + 3 + 5 + 11 \quad \text{and} \quad (13 + 7 + 1) - (3 + 5 + 11) = 2
     \]
   - If we remove \(11\):
     \[
     13 + 5 + 1 = 3 + 7 + 9 \quad \text{and} \quad (1 + 3 + 7 + 9) - (13 + 5) = 2
     \]
   - If we remove \(13\):
     \[
     11 + 7 = 1 + 3 + 5 + 9 \quad \text{and} \quad (11 + 7 + 1) - (3 + 5 + 9) = 2
     \]

2. **Inductive Step:**
   - Assume the result is true for some odd \( n \geq 7 \). We need to show it holds for \( n + 2 \).
   - Consider the set \(\{1, 3, 5, \ldots, 2n-1, 2n+1, 2n+3\}\).

   If we remove any coin \( w_i \) where \( 1 \leq w_i \leq 2n-1 \), by the inductive hypothesis, there exist partitions \( A \cup B \) and \( C \cup D \) such that:
   \[
   \sum_{x \in A} x = \sum_{x \in B} x \quad \text{and} \quad \sum_{x \in C} x = 2 + \sum_{x \in D} x
   \]

   Define new sets:
   \[
   A' = D \cup \{2n+3\}, \quad B' = C \cup \{2n+1\}
   \]
   Then:
   \[
   \sum_{x \in A'} x = \sum_{x \in B'} x
   \]

   Similarly, define:
   \[
   C' = A \cup \{2n+3\}, \quad D' = B \cup \{2n+1\}
   \]
   Then:
   \[
   \sum_{x \in C'} x - \sum_{x \in D'} x = 2
   \]

3. **Special Cases for \( n = 4k-1 \) and \( n = 4k+1 \):**
   - For \( n = 4k-1 \):
     - If we remove \( 2n+3 \), we can form \( 2k \) pairs, each summing to \( 2n+2 \). Take \( k \) pairs to form \( A \) and include \(\{1, 2n+1\}\) in \( A \). The rest form \( B \).
     - If we remove \( 2n+1 \), adjust the sets \( A \) and \( B \) accordingly to maintain the required sums.

   - For \( n = 4k+1 \):
     - If we remove \( 2n+3 \), form pairs and adjust sets \( C \) and \( D \) to maintain the required sums.
     - If we remove \( 2n+1 \), adjust the sets \( A \) and \( B \) accordingly.

Thus, the solution holds for all odd \( n \geq 7 \).

The final answer is \( \boxed{ n } \) for all odd \( n \geq 7 \).