Let $(M,g)$ be a closed(oriented) Riemannian 4 manifold. It is well-known that, if $scal^g\geq0$ and not identically zero, then $M$ admits a PSC metric by conformal transformation. 


**Q** Is $T^4$ the only oriented closed 4 manifold, which admits metric with flat scalar curvature but not PSC metric? 



Flat scalar curvature on 4 manifold
No : K3 surfaces admit zero scalar curvature metrics (in fact even Ricci flat ones) but no metric with positive scalar curvature. See for instance the lecture notes by Gromov ([Four lectures on scalar curvature](https://www.ihes.fr/%7Egromov/wp-content/uploads/2019/04/scalar-lectures-IHES-2019.pdf)) p. 18.


This comes from the Weitzenbock formula for the Dirac operator and the index theorem. It is attributed to Lichnerowicz.


As pointed out by Danny Ruberman in comment, any flat four manifold gives an example, although this was proved later by Schoen and Yau.


Another well known observation is that any scalar flat manifold which does not admit a PSC metric is Ricci flat. This was proved by Kazhdan if I recall but I can't remember his proof. An alternative proof uses Ricci flow :


Along Ricci flow we have :
$$ \frac{\partial \text{s}\_{g(t)}}{\partial t}=\Delta\_{g(t)}\text{s}\_{g(t)}+2|\text{ric}\_{g(t)}|^2$$


If one starts the Ricci flow with a scalar flat manifold $(M,g\_0)$, the maximum principle will show it will either either get positive scalar curvature for $t>0$ or stay scalar flat. If $M$ has no PSC metric, we are in the second case. Now using $\text{s}\_{g(t)}=0$ in the evolution equation above we have $|\text{ric}\_{g(t)}|^2=0$ and (since $g(t)$ is Ricci flat) $g(t)=g\_0$.


