As Fedor Petrov mentions in the comments, a necessary and sufficient condition is that each vertex has even degree.  Here is a proof.  

Let $T^*$ be the dual graph of the triangulation.  That is, the vertices of $T$ are the faces of the triangulation, and two faces are adjacent if they share an edge.  Rephrased, your question is asking when $T^*$ is bipartite.  So, it suffices to prove that a planar graph $G$ is bipartite if and only if the dual graph $G^*$ is Eulerian (all vertices have even degree). For one direction, if $G$ is bipartite, then all cycles of $G$ are even.  In particular, all facial cycles are even, and hence $G^*$ is Eulerian.  For the other direction, suppose that the dual graph $G^*$ is Eulerian.  Thus, each face of $G$ is even.  Since the cycle space of $G$ is generated by the facial cycles of $G$ (take the symmetric difference of all faces inside the cycle), this implies that *every* cycle of $G$ is even.  Hence $G$ is bipartite.