Natural $n \geq 2$ is given. Group of people calls $n-compact$, if for any men from group, we can found $n$ people (without he), each two of there are familiar. 
Find maximum $N$ such that for any $n-compact$ group, consisting $N$ people contains subgroup from $n+1$ people, each of two of there are familiar.
1. **Understanding the Problem:**
   We need to find the maximum number \( N \) such that any \( n \)-compact group of \( N \) people contains a subgroup of \( n+1 \) people, each of whom knows each other. A group is \( n \)-compact if for any person in the group, there exist \( n \) people who all know each other.

2. **Restating the Problem in Graph Theory Terms:**
   - Let \( G \) be a graph where vertices represent people and edges represent familiarity (knowing each other).
   - The group is \( n \)-compact if for any vertex \( v \), there exists a clique of size \( n \) in the neighborhood of \( v \) (i.e., among the vertices adjacent to \( v \)).
   - We need to find the maximum \( N \) such that any \( n \)-compact graph with \( N \) vertices contains a clique of size \( n+1 \).

3. **Using Turán's Theorem:**
   Turán's theorem states that the maximum number of edges in a graph on \( N \) vertices that does not contain a \( K_{r+1} \) (a complete graph on \( r+1 \) vertices) is given by:
   \[
   \text{ex}(N, K_{r+1}) = \left(1 - \frac{1}{r}\right) \frac{N^2}{2}
   \]
   where \( r \) is the size of the largest clique minus one.

4. **Applying Turán's Theorem:**
   - We need to ensure that our graph contains a \( K_{n+1} \).
   - If \( N \) is too large, the graph might not be \( n \)-compact.
   - We need to find the threshold where the graph must contain a \( K_{n+1} \).

5. **Finding the Maximum \( N \):**
   - For a graph to be \( n \)-compact, each vertex must have a neighborhood that contains a \( K_n \).
   - If \( N \leq 2n \), then by the pigeonhole principle, there must be a vertex with a neighborhood of size at least \( n \), ensuring a \( K_n \) in the neighborhood.
   - If \( N = 2n+1 \), we need to check if this guarantees a \( K_{n+1} \).

6. **Verification:**
   - Consider a graph with \( N = 2n+1 \) vertices.
   - If the graph is \( n \)-compact, each vertex has a neighborhood containing a \( K_n \).
   - By the Erdős–Stone theorem, a graph with \( 2n+1 \) vertices and sufficiently many edges will contain a \( K_{n+1} \).

7. **Conclusion:**
   - The maximum \( N \) such that any \( n \)-compact group of \( N \) people contains a subgroup of \( n+1 \) people, each of whom knows each other, is \( 2n+1 \).

The final answer is \( \boxed{ 2n+1 } \).