A measure $\mu\in\mathcal{M}(X)$ is called strictly positive if every non-empty open subset of $X$ has strictly positive measure. Also $\mu\in\mathcal{M}(X)$ is $f$-invariant, if for every measurable set $A$, $\mu(f^{-1}(A))= \mu(A)$ ($f:X\to X$ is a dynamical system).
It is known that if $f:X\to X$ is a homeomorphism on compact metric space, then there is $\mu\in\mathcal{M}(X)$ such that $\mu$ is $f$-invariant measure. Also there is a strictly positive measure on every polish space without isolated point.
In my research I work with a homeomorphism $f:X\to X$ on compact metric space without isolated point. Is there a strictly positive measure $\mu\in\mathcal{M}(X)$ such that $\mu$ is $f$-invariant measure?
Then the answer is no (if you want a finite measure; otherwise, the answer is yes: take the counting measure), not even for merely finitely additive measures.
To see this, consider the action of $\mathbf Z$ on the extended real line $[-\infty,\infty]$ by translation (or even just the one-point compactification of $\mathbf Z$). A strictly positive measure would necessarily be infinite, simply because each open interval has infinitely many disjoint translates. In fact, it would not even be locally finite: any neighbourhood of $\pm\infty$ contains infinitely many disjoint translates of every interval.
Note that if you flow is minimal (meaning, every orbit is dense), then actually every nonzero invariant measure is strictly positive (so the answer is yes in this case, for $\mathbf Z$-flows anyway, since invariant measures do exist). You do not even need to assume metrisability, but let us do so for simplicity. Take any ball $B$ centered at $x$ of radius $r$. Then the union of all translates of $B$ is the whole $X$: indeed, for any $y\in X$, we can find some $x'$ in the orbit of $x$ such that $d(x',y)<r$, and then $y$ is in the corresponding ball $B'$. But then by compactness, some $M<\infty$ translates of $B$ cover all of $X$. It follows that $\mu(X)\leq M\mu(B)$, whence $\mu(B)\geq \mu(X)/M>0$.
It follows that if $X$ is the finite disjoint union of its minimal subflows, then it supports a strictly positive invariant probability measure.
I suspect that as soon as you have strict containment between two closed orbits, you will have no strictly positive (finite) invariant measures.