Is there a set $S$ of positive integers such that a number is in $S$ if and only if it is the sum of two distinct members of $S$ or a sum of two distinct positive integers not in $S$?
To determine if there exists a set \( S \) of positive integers such that a number is in \( S \) if and only if it is the sum of two distinct members of \( S \) or a sum of two distinct positive integers not in \( S \), we can construct such a set and verify its properties.

1. **Define the sets \( S \) and \( T \):**
   \[
   S = \{3, 5, 6, 8, 9, 11, 12, 13, \ldots\}
   \]
   \[
   T = \mathbb{Z}^+ \setminus S = \{1, 2, 4, 7, 10\}
   \]

2. **Verify that elements in \( S \) are sums of two distinct elements in \( T \):**
   - \( 3 = 1 + 2 \)
   - \( 5 = 1 + 4 \)
   - \( 6 = 2 + 4 \)
   - \( 8 = 1 + 7 \)
   - \( 9 = 2 + 7 \)
   - \( 11 = 1 + 10 \) or \( 4 + 7 \)
   - \( 12 = 2 + 10 \)
   - \( 13 = 3 + 10 \) (both in \( S \))

3. **Verify that elements in \( S \) are sums of two distinct elements in \( S \):**
   - For \( a \geq 13 \), we can use induction to show that \( a \) is a sum of two distinct members of \( S \). For example:
     - \( 13 = 5 + 8 \)
     - \( 14 = 6 + 8 \)
     - \( 15 = 6 + 9 \)
     - \( 16 = 8 + 8 \) (but we need distinct members, so consider \( 5 + 11 \))
     - \( 17 = 8 + 9 \)
     - \( 18 = 9 + 9 \) (but we need distinct members, so consider \( 6 + 12 \))
     - And so on.

4. **Verify that sums of two distinct elements in \( T \) are in \( S \):**
   - \( 1 + 2 = 3 \)
   - \( 1 + 4 = 5 \)
   - \( 1 + 7 = 8 \)
   - \( 1 + 10 = 11 \)
   - \( 2 + 4 = 6 \)
   - \( 2 + 7 = 9 \)
   - \( 2 + 10 = 12 \)
   - \( 4 + 7 = 11 \)
   - \( 4 + 10 = 14 \)
   - \( 7 + 10 = 17 \)

5. **Verify that sums of two distinct elements in \( S \) are in \( S \):**
   - If \( a, b \in S \) and \( a + b \geq 11 \), then \( a + b \in S \).
   - If \( a + b \leq 10 \), then \( (a, b) \in \{(3, 5), (3, 6)\} \) and their sums are in \( S \):
     - \( 3 + 5 = 8 \)
     - \( 3 + 6 = 9 \)

Thus, the set \( S \) satisfies the given condition.

The final answer is \( \boxed{ S = \{3, 5, 6, 8, 9, 11, 12, 13, \ldots\} } \)