Let $M$ be a finite sum of numbers, such that among any three of its elements there are two whose sum belongs to $M$.  Find the greatest possible number of elements of $M$.
1. **Define the set \( M \) and the condition:**
   Let \( M \) be a finite set of numbers such that among any three of its elements, there are two whose sum belongs to \( M \).

2. **Establish the upper bound:**
   We need to show that \( M \) can have at most 7 elements. To see this, consider the set \( M = \{-3, -2, -1, 0, 1, 2, 3\} \).

3. **Verify the condition for the example set:**
   - For any three elements \( a, b, c \in M \), we need to check if \( a + b \in M \), \( a + c \in M \), or \( b + c \in M \).
   - For example, if we take \( a = -3, b = -2, c = -1 \):
     - \( a + b = -5 \notin M \)
     - \( a + c = -4 \notin M \)
     - \( b + c = -3 \in M \)
   - This shows that the condition is satisfied for this set.

4. **Consider the general case:**
   Suppose \( M \) has \( \geq 4 \) elements of the same sign. Without loss of generality, let \( a_1, a_2, a_3, a_4 \) be those elements with the greatest absolute values, and assume \( |a_1| > |a_2| > |a_3| > |a_4| \).

5. **Derive a contradiction:**
   - By selecting \( a_1, a_2, a_3 \), we have \( a_2 + a_3 = a_1 \).
   - By selecting \( a_1, a_2, a_4 \), we get \( a_2 + a_4 = a_1 \).
   - This leads to a contradiction because \( a_2 + a_3 \neq a_2 + a_4 \) unless \( a_3 = a_4 \), which is not possible given the distinct absolute values.

6. **Conclude the maximum number of elements:**
   - Therefore, \( M \) can have at most 3 positive elements and at most 3 negative elements.
   - Including 0, the total number of elements in \( M \) is at most \( 3 + 3 + 1 = 7 \).

The final answer is \( \boxed{ 7 } \).