In a simple graph $G$, we call $t$ pairwise adjacent vertices a $t$[i]-clique[/i]. If a vertex is connected with all other vertices in the graph, we call it a [i]central[/i] vertex. Given are two integers $n,k$ such that $\dfrac {3}{2} \leq \dfrac{1}{2} n < k < n$. Let $G$ be a graph on $n$ vertices such that
[b](1)[/b] $G$ does not contain a $(k+1)$-[i]clique[/i];
[b](2)[/b] if we add an arbitrary edge to $G$, that creates a $(k+1)$-[i]clique[/i].
Find the least possible number of [i]central[/i] vertices in $G$.
1. **Understanding the Problem:**
   We are given a graph \( G \) with \( n \) vertices and two integers \( n \) and \( k \) such that \(\frac{3}{2} \leq \frac{1}{2} n < k < n\). The graph \( G \) satisfies two conditions:
   - \( G \) does not contain a \((k+1)\)-clique.
   - Adding any edge to \( G \) creates a \((k+1)\)-clique.
   
   We need to find the least possible number of central vertices in \( G \).

2. **Initial Claim:**
   The solution claims that the answer is \( \boxed{2k - n} \). This can be derived by considering the Turan graph \( T_{n, k} \).

3. **Turan's Theorem:**
   Turan's Theorem states that the maximum number of edges in a graph on \( n \) vertices that does not contain a \((k+1)\)-clique is given by the Turan graph \( T_{n, k} \), which is a \( k \)-partite graph with parts as equal as possible.

4. **Minimal Number of Central Vertices:**
   Consider the graph with the minimal number of central vertices. We claim that there do not exist three vertices \( u, v, w \) in the graph such that \( v \) is adjacent to \( w \) and \( u \) is not adjacent to \( v \) and not adjacent to \( w \). Assume for the sake of contradiction that such vertices exist.

5. **Central Vertex \( a \):**
   Consider a central vertex \( a \). We claim that any \((k+1)\)-clique that exists after an edge is added between vertices \( u \) and \( v \) contains \( a \). Assume the contrary, then there is a \( k-1 \)-clique all of whose vertices are connected to \( u \) and \( v \) and none of whose vertices are \( a \). But then the vertices of that \( k-1 \)-clique and \( a \) and \( u \) form a \((k+1)\)-clique, which is a contradiction.

6. **Modifying the Graph:**
   Now, delete the edge between \( a \) and \( v \) and add an edge between \( v \) and \( u \). We claim that this new graph satisfies all the properties outlined in the problem and has fewer central vertices than the graph we started with. Since \( a \) is no longer a central vertex and since clearly no vertex became a new central vertex in the new graph, it suffices to check that the new graph contains no \((k+1)\)-clique and adding any edge to the new graph results in a graph with a \((k+1)\)-clique.

7. **Verification:**
   Since all \((k+1)\)-cliques formed by adding an edge between \( u \) and \( v \) contained \( a \) and we deleted the edge between \( a \) and \( v \), we know that the new graph contains no \((k+1)\)-clique. Also, if we add an edge between \( a \) and \( v \) in the new graph, we obtain the same graph as if we had added an edge between \( u \) and \( v \) in the original graph, which by definition yields a graph containing a \((k+1)\)-clique. Therefore, the new graph contradicts the minimality of our original graph.

8. **Conclusion:**
   This means that "not being adjacent" is an equivalence relation between vertices, which means that our graph is complete and partite. Since it must be at least \( k \)-partite, we easily obtain the desired result.