The Wikipedia article on scalar curvature says that
If $X$ is a compact Kähler manifold of complex dimension 2 which is not rational or ruled, then $X$ (as a smooth 4-manifold) has no Riemannian metric with positive scalar curvature.
Question: Let $M$ be homeomorphic to $S^4$. Give $M$ some differentiable structure. Does $M$ always admit a metric of positive scalar curvature?
I am not familiar with complex geometry, so I don't know if $M$ satisfies the above conditions.
If we assume $M$ has the standard differentiable structure of $S^4$, then the standard metric of $S^4$ has positive scalar curvature. However, I would like to know if the answer to my question is true for purely topological reasons, i.e. without assuming a differentiable structure on $M$ a priori.
It is an open problem as to whether or not $S^4$ admits a non-standard smooth structure (such a smooth structure is called an exotic smooth structure). This problem is known as the four-dimensional smooth Poincaré conjecture.
No matter which smooth structure $S^4$ is endowed with, it cannot be given a Kähler complex structure as a compact Kähler manifold $M$ has $b_2(M) = \dim H^2(M) > 0$. In fact, no smooth $S^4$ (standard or exotic) can even admit an almost complex structure.
I don't know the answer to your actual question, but I wouldn't be surprised if it is open.