Let $n\ge 4$ be a positive integer and let $M$ be a set of $n$ points in the plane, where no three points are collinear and not all of the $n$ points being concyclic. Find all real functions $f:M\to\mathbb{R}$ such that for any circle $\mathcal{C}$ containing at least three points from $M$, the following equality holds:
\[\sum_{P\in\mathcal{C}\cap M} f(P)=0\]
[i]Dorel Mihet[/i]
1. **Define Interesting Circles:**
   Call the circles which contain at least three of the points from \( M \) interesting. The problem's conditions tell us that every triplet of the points belongs to exactly one interesting circle.

2. **Lemma:**
   \[
   \sum_{P \in M} f(P) = 0
   \]

   **Proof:**
   For a triplet of points \( T = \{P, Q, R\} \), let \( f(T) = f(P) + f(Q) + f(R) \). Consider \( \sum_{T \subset M} f(T) \), with \( T \) ranging over every possible triplet.

3. **Counting Triplets:**
   On one hand, every point of \( M \) is included in exactly \( \binom{n-1}{2} \) triplets, so:
   \[
   \sum_{T \subset M} f(T) = \binom{n-1}{2} \sum_{P \in M} f(P)
   \]

4. **Triplets on Interesting Circles:**
   On the other hand, suppose we choose an interesting circle \( \mathcal{C} \) and count all the triplets which are entirely on it. If the circle has \( k \) points from \( M \), each point will be included in \( \binom{k-1}{2} \) triplets; thus:
   \[
   \sum_{T \subset M \cap \mathcal{C}} f(T) = \binom{k-1}{2} \sum_{P \in M \cap \mathcal{C}} f(P) = 0
   \]
   because \( \sum_{P \in M \cap \mathcal{C}} f(P) = 0 \) by the problem's condition.

5. **Summing Over All Interesting Circles:**
   Now we simply sum over all interesting circles; since each triplet is included in exactly one interesting circle, we have:
   \[
   \sum_{T \subset M} f(T) = 0
   \]
   Combining these two equations proves the lemma:
   \[
   \binom{n-1}{2} \sum_{P \in M} f(P) = 0 \implies \sum_{P \in M} f(P) = 0
   \]

6. **Analyzing Points on Circles:**
   Let \( A \) and \( B \) be two arbitrary points and let \( S \) be the set of all interesting circles which pass through both of them. Not all of the points of \( M \) are concyclic, hence \( |S| > 1 \). Also, every point other than \( A \) and \( B \) which is in \( M \) is in exactly one circle of \( S \), so:
   \[
   0 = \sum_{\mathcal{C} \in S} \sum_{P \in \mathcal{C} \cap M} f(P) = |S|(f(A) + f(B)) + \sum_{P \in M \setminus \{A, B\}} f(P)
   \]
   Applying the lemma, we have:
   \[
   0 = (|S| - 1)(f(A) + f(B))
   \]
   Since \( |S| > 1 \), we know:
   \[
   f(A) + f(B) = 0
   \]

7. **Conclusion:**
   For three arbitrary points \( A, B, C \), we have:
   \[
   2f(A) = (f(A) + f(B)) + (f(A) + f(C)) - (f(B) + f(C)) = 0
   \]
   So the only function that works is the zero function:
   \[
   f(P) = 0 \quad \text{for all} \quad P \in M
   \]

The final answer is \( \boxed{ f(P) = 0 } \) for all \( P \in M \).