Find all integers $n \geq 3$ with the following property: there exist $n$ distinct points on the plane such that each point is the circumcentre of a triangle formed by 3 of the points.
To solve the problem, we need to find all integers \( n \geq 3 \) such that there exist \( n \) distinct points on the plane where each point is the circumcenter of a triangle formed by three of the points. We will analyze different cases based on the value of \( n \).

1. **Case \( n \geq 7 \)**:
   - Consider a regular hexagon \( A_1A_2A_3A_4A_5A_6 \) and its center \( O \). This configuration works because each vertex of the hexagon and the center can serve as the circumcenter of triangles formed by other vertices.
   - For \( n \in \{8, 9, 10, 11, 12, 13\} \), we can add points \( O_i \), which are the circumcenters of triangles \( \triangle OA_iA_{i+1} \). For example, if \( n = 11 \), we add \( O_1, O_2, O_3, O_4 \).
   - By induction, for each \( n \geq 7 \), we can construct a configuration. If \( n = 7k + r \) with \( 0 \leq r \leq 6 \), consider \( k-1 \) disjoint hexagons with their centers and another one adding their \( O_1, O_2, \ldots, O_r \).

2. **Case \( n = 3 \)**:
   - With only 3 points, say \( A, B, C \), one of the points must be the circumcenter of \( \triangle ABC \). This is not possible because the circumcenter of a triangle cannot be one of its vertices.

3. **Case \( n = 4 \)**:
   - Let the points be \( A_1, A_2, A_3, A_4 \). Each \( A_i \) must be the circumcenter of \( \triangle A_{i+1}A_{i+2}A_{i+3} \). This implies all edges \( A_iA_j \) must be equal, making \( \triangle A_1A_2A_3 \) and \( \triangle A_1A_2A_4 \) equilateral. However, \( \triangle A_3A_2A_4 \) cannot be equilateral since \( \angle A_3A_2A_4 = 120^\circ \).

4. **Case \( n = 5 \)**:
   - We denote \( i \rightarrow jkl \) if \( A_i \) is the circumcenter of \( \triangle A_jA_kA_l \). Consider \( 4 \rightarrow 123 \) and \( 5 \rightarrow 234 \).
   - \( A_1 \) cannot be the circumcenter of \( \triangle A_2A_3A_4 \) from the \( n = 3 \) case.
   - **Case 1**: \( 1 \rightarrow 245 \) or \( 1 \rightarrow 345 \). Without loss of generality, assume \( 1 \rightarrow 245 \). Then \( 3 \rightarrow 145 \) or \( 3 \rightarrow 125 \). Both cases lead to contradictions because they imply multiple equilateral triangles sharing sides, which is impossible.
   - **Case 2**: \( 1 \rightarrow 253 \). Here, \( A_1, A_4, \) and \( A_5 \) are collinear on the perpendicular bisector of \( A_2A_3 \). Using Cartesian coordinates, we can show that the distances do not satisfy the required conditions.

5. **Case \( n = 6 \)**:
   - This case is more complex and involves multiple sub-cases. However, it can be shown that no configuration exists where each point is the circumcenter of a triangle formed by three other points.

Conclusion:
From the analysis, we conclude that the only possible values for \( n \) are \( n \geq 7 \).

The final answer is \( \boxed{ n \geq 7 } \).