View the configuration as a simple graph. Note that if two vertices have no common neighbors then we can draw an edge between them without getting a $K_3$ subgraph. So we need every vertices of points to have a common neighbor. We can achieve this with one vertex connected to all $2013$ other points, which is $2013$ segments total. If there are any less than $2013$ edges, then the graph isn't connected, i.e. there are two vertices that do not have a path between them. Then it is possible to draw an edge between those two vertices, contradiction. In general, the minimum number of edges required for $n$ points is $\boxed{n-1}$.