There are some people in a meeting; each doesn't know at least 56 others, and for any pair, there exist a third one who knows both of them. Can the number of people be 65?
To determine if the number of people in the meeting can be 65, we need to analyze the given conditions and use graph theory concepts. Let's break down the problem step by step.

1. **Graph Representation**:
   - Represent the people in the meeting as vertices of a graph \( G \).
   - An edge between two vertices indicates that the corresponding people know each other.
   - Each person (vertex) does not know at least 56 others, implying that each vertex has a degree of at most \( 64 - 56 = 8 \).

2. **Triangle-Free Condition**:
   - For any pair of vertices, there exists a third vertex that knows both of them. This implies that the graph is triangle-free (no three vertices form a triangle).

3. **Counting Closed Walks of Length 7**:
   - We need to count the number of closed walks of length 7, denoted as \( f(7) \).
   - A closed walk of length 7 is a sequence of vertices \( v_0, v_1, v_2, \ldots, v_7 \) where \( v_i \) and \( v_{i+1} \) are connected and \( v_0 = v_7 \).

4. **Divisibility by 7**:
   - If \( v_0, v_1, v_2, \ldots, v_0 \) is a closed walk of length 7, then so are \( v_1, v_2, v_3, \ldots, v_1 \), \( v_2, v_3, v_4, \ldots, v_2 \), and so on.
   - Therefore, \( f(7) \) is divisible by 7.

5. **Counting Closed Walks with a Fixed Vertex \( v_0 \)**:
   - Fix a vertex \( v_0 \) and count the number of closed walks \( v_0, v_1, v_2, \ldots, v_0 \).
   - If \( v_5 = v_0 \), the number of such walks is \( 8f(5) \).
   - If \( v_5 \neq v_0 \), the number of such walks is \( 8^5 - f(5) - X \), where \( X \) is the number of times \( v_5 \) and \( v_0 \) are connected.

6. **Finding \( X \)**:
   - If \( v_5 \) is connected to \( v_0 \), then \( v_4 \) is not a neighbor of \( v_0 \) (to avoid forming a triangle).
   - Let \( Y \) be the number of walks where \( v_4 \) is connected to \( v_0 \), and \( Z \) be the number of walks where \( v_4 \) is not connected to \( v_0 \).
   - \( Z = 8^4 - Y \).
   - \( Y = 8 \times 7^2 \) because from \( v_0 \) we have 8 possibilities for \( v_1 \), 7 possibilities for \( v_2 \) (excluding \( v_0 \)), 7 possibilities for \( v_3 \), and a unique vertex connected to both \( v_0 \) and \( v_3 \).
   - Therefore, \( Z = 8^4 - 8 \times 7^2 \).

7. **Finding \( T \)**:
   - \( T \) is the number of walks where \( v_4 = v_0 \).
   - \( T = 8 \times 15 \).

8. **Combining Results**:
   - \( X = Z + T = 8^4 - 8 \times 7^2 + 8 \times 15 \).
   - \( f(7) = 8f(5) + 8^5 - f(5) - X = 7f(5) + 8^5 - (8^4 - 8 \times 7^2 + 8 \times 15) \equiv -1 \pmod{7} \).

9. **Summing Over All Vertices**:
   - Summing over all vertices, we get \( -65 \equiv 0 \pmod{7} \), which is a contradiction.

Therefore, the number of people in the meeting cannot be 65.

The final answer is \( \boxed{\text{False}} \).