Given a continuous (surjective) map of a compact metric space, the Kryloff-Bogolioubuff argument shows that there is at least $T$ invariant measure and hence one or uncountably many measures for which $T$ is ergodic. Is every Borel measure ergodic (and invariant) for some continuous map?
An ergodic invariant measure associated to a periodic orbit is uniform over that orbit. Therefore, a finitely supported measure that is not uniform over its support is not an ergodic invariant measure for any measurable transformation.
Update: I suspect that the answer remains negative even if you assume the measure to be non-atomic, but I don't know a counter-example. However, if you do not require the map to be continuous (but merely measurable), then the answer becomes positive: every non-atomic probability measure on a compact metric space is ergodic for some measurable transformation.
Namely, if $X$ is a complete separable metric space and $\mu$ is a non-atomic Borel probability measure on $X$, then the measure space $(X,\mu)$ is isomorphic mod $0$ to the unit interval with the Lebesgue measure. Take any map on the unit interval for which the Lebesgue measure is ergodic (say an irrational rotation) and lift it to $X$.
Any constant map is continuous and the only invariant sets are $\emptyset$ and the whole space. Hence, any measure is ergodic with respect to a constant map.
But I suspect that you meant invariant ergodic measures. Nice question but that is a whole different matter and I am strictly answering your question. However, note that ergodic theory doesn't start with invariant measures. For example, a measure is ergodic if and only if any function that is invariant almost everywhere is in fact constant almost everywhere. This does not require the measure to be invariant!
Just a little correction: Krylov-Bogolubov's theorem requires a compact metric space.