Let $n$ points be given arbitrarily in the plane, no three of them collinear. Let us draw segments between pairs of these points. What is the minimum number of segments that can be colored red in such a way that among any four points, three of them are connected by segments that form a red triangle?
1. **Understanding the Problem:**
   We are given \( n \) points in the plane with no three points collinear. We need to determine the minimum number of segments that must be colored red such that among any four points, three of them form a red triangle.

2. **Initial Assumptions:**
   Assume the edge \( ab \) is not red. This implies that for any four points \(\{a, b, x, y\}\), there must be a red triangle among them.

3. **Implications of the Assumption:**
   Since \( ab \) is not red, the edges \( xy \) must be red. Additionally, either \( ax \) and \( ay \) must be red, or \( bx \) and \( by \) must be red. This ensures that among any four points, three of them form a red triangle.

4. **Red Subgraph:**
   This implies that the complete graph \( K_n \) minus the edge \( ab \) (denoted as \( K_n - \{a, b\} \)) must be red. Let \( A \) be the set of vertices \( x \) such that \( ax \) is red, and \( B \) be the set of vertices \( y \) such that \( by \) is red. It follows that \( A \cup B = K_n \setminus \{a, b\} \).

5. **Contradiction and Conclusion:**
   If we could find \( x \in A \setminus B \) and \( y \in B \setminus A \), then \(\{a, b, x, y\}\) would not satisfy the condition, leading to a contradiction. Therefore, \( B \setminus A = \emptyset \), implying \( A = K_n \setminus \{a, b\} \). Thus, \( K_n - \{b\} = K_{n-1} \) is red.

6. **Final Calculation:**
   The minimum number of red edges required is the number of edges in \( K_{n-1} \), which is given by:
   \[
   m(n) = \binom{n-1}{2} = \frac{(n-1)(n-2)}{2}
   \]

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