Given a graph with $n \geq 4$ vertices. It is known that for any two of vertices there is a vertex connected with none of these two vertices.
Find the greatest possible number of the edges in the graph.
1. **Complement Graph Analysis**:
   - Let \( G \) be the complement of the given graph. In \( G \), for every pair of vertices \( A \) and \( B \), there exists a vertex \( C \) that is connected to both \( A \) and \( B \).
   - Let \( E \) denote the number of edges in \( G \).

2. **Minimum Degree Condition**:
   - Since for every pair of vertices in \( G \) there exists a vertex connected to both, it implies that \( \delta(G) \geq 1 \) (minimum degree of \( G \) is at least 1).

3. **Counting Triples**:
   - We count the number of triples \( (A, B, C) \) where \( A, B, C \) are vertices in \( G \) and \( C \) is connected to both \( A \) and \( B \).
   - Fixing \( C \), the number of such triples is \( \sum \binom{d_i}{2} \), where \( d_i \) is the degree of vertex \( i \) in \( G \).

4. **Inequality for Triples**:
   - The total number of such triples must be at least \( \binom{n}{2} \) because for every pair of vertices, there is at least one vertex connected to both.
   - Therefore, \( \sum \binom{d_i}{2} \geq \binom{n}{2} \).

5. **Combining Degrees**:
   - For any two integers \( a \) and \( b \) both at least 1, we have \( \binom{a}{2} + \binom{b}{2} \leq \binom{a+b-2}{2} + 1 \).
   - Using this inequality by combining the vertices in descending order of their degree, we get:
     \[
     \sum \binom{d_i}{2} \leq n-1 + \binom{2E - 2(n-1)}{2}
     \]

6. **Applying the Inequality**:
   - Using the above inequality, we get:
     \[
     \binom{n-1}{2} \leq \binom{2E - 2(n-1)}{2}
     \]
   - Simplifying this, we find:
     \[
     E \geq \frac{3(n-1)}{2}
     \]

7. **Upper Bound on Edges**:
   - Examining the given graph, the number of edges in the complement graph \( G \) is:
     \[
     E \leq \frac{n(n-1)}{2} - \frac{3(n-1)}{2}
     \]
   - Simplifying this, we get the desired upper bound:
     \[
     E \leq \frac{(n-1)(n-3)}{2}
     \]

The final answer is \(\boxed{\frac{(n-1)(n-3)}{2}}\).