Thus, considering the graph with vertices the $2014$ airports, and edges between pairs of airports being the flights, as described, we are to prove there exists no Hamiltonian path. The solution is remarkably fast. Consider the convex hull of the airports; it will have at least three vertices. For each vertex of the convex hull it is trivial there exists exactly one edge incident at it (by continuity, rotating a line vith the vertex as pivot, there will be a unique moment when the line will separate the country). So there are at least three vertices of degree $1$, and that prohibits the existence of a Hamiltonian path, where all but maybe two of the vertices (the ends of the path) must have degree at least $2$.