Given $2014$ points in the plane, no three of which are collinear, what is the minimum number of line segments that can be drawn connecting pairs of points in such a way that adding a single additional line segment of the same sort will always produce a triangle of three connected points?
1. **Interpret the problem in terms of graph theory:**
   - We are given 2014 points in the plane, no three of which are collinear.
   - We need to find the minimum number of line segments (edges) such that adding any additional line segment will always form a triangle (a $K_3$ subgraph).

2. **Graph Theory Translation:**
   - Consider the points as vertices of a graph.
   - We need to ensure that every pair of vertices has a common neighbor, which implies that adding any new edge will create a triangle.

3. **Constructing the Graph:**
   - To achieve the condition where adding any new edge forms a triangle, we can use a star graph configuration.
   - In a star graph, one central vertex is connected to all other vertices. This ensures that every pair of vertices has a common neighbor (the central vertex).

4. **Counting the Edges:**
   - In a star graph with 2014 vertices, the central vertex is connected to the remaining 2013 vertices.
   - Therefore, the number of edges in this configuration is 2013.

5. **Verification:**
   - If there are fewer than 2013 edges, the graph is not connected, meaning there exist at least two vertices with no path between them.
   - Adding an edge between any two such vertices would not necessarily form a triangle, contradicting the problem's requirement.

6. **Generalization:**
   - For \( n \) points, the minimum number of edges required to ensure that adding any additional edge forms a triangle is \( n-1 \).

Conclusion:
\[
\boxed{2013}
\]