We have a complete graph with $n$ vertices. We have to color the vertices and the edges in a way such that: no two edges pointing to the same vertice are of the same color; a vertice and an edge pointing him are coloured in a different way. What is the minimum number of colors we need?
1. **Claim**: The smallest number of colors required is \( \boxed{n} \).

2. **Construction**: 
   - Color each vertex \( v_i \) with color \( i \) for \( i = 0, 1, 2, \ldots, n-1 \).
   - Color the edge \( v_i v_j \) with color \( (i + j) \mod n \).

3. **Verification**:
   - **Vertex-Edge Coloring**: For any vertex \( v_i \), the edge \( v_i v_j \) is colored with \( (i + j) \mod n \). Since \( i \) and \( (i + j) \mod n \) are different (as \( j \neq 0 \)), the vertex and the edge pointing to it are colored differently.
   - **Edge-Edge Coloring**: Consider any vertex \( v_i \). The edges \( v_i v_j \) and \( v_i v_{j'} \) are colored with \( (i + j) \mod n \) and \( (i + j') \mod n \) respectively. If these two edges have the same color, then:
     \[
     (i + j) \mod n = (i + j') \mod n
     \]
     This implies:
     \[
     i + j \equiv i + j' \pmod{n}
     \]
     Simplifying, we get:
     \[
     j \equiv j' \pmod{n}
     \]
     Since \( j \) and \( j' \) are both in the range \( 0, 1, 2, \ldots, n-1 \), this means \( j = j' \). Hence, no two edges pointing to the same vertex are of the same color.

4. **Conclusion**: The minimum number of colors required to color the vertices and edges of a complete graph with \( n \) vertices, such that no two edges pointing to the same vertex are of the same color and a vertex and an edge pointing to it are colored differently, is \( \boxed{n} \).