On the Domino Train

Consider a graph with vertices labeled $1, 2, \ldots , 2016$ . Then an edge connecting $a$ and $b$ on the graph corresponds to a domino with numbers $[a|b]$ . A train is simply drawing edges from vertex to vertex without lifting up one's pencil which is an Eulerian path, and using up all the dominos corresponds to drawing a complete graph $K_{2016}$ plus a loop at each vertex that starts and ends there (corresponding to the dominoes with the same number on both ends).

Note that every single vertex of this graph has degree $2017$ which is odd. Because an Eulerian path can pass through every edge of at most $2$ vertices with odd degree, each Eulerian path will only get rid of $2$ odd-degree vertices (all the other vertices it passes through will only have an even number less edges untraveled, which doesn't affect its odd-degree status). Thus we must have at least $2016\div 2=\boxed{1008}$ different Eulerian paths, or trains.