Conformally flat manifold with zero scalar I would like to ask the following : Is there any example of a compact conformally flat Riemannian manifold $(M^n,g)$ with $n\geq 4$ which is not flat and has zero scalar curvature? 
 A: Take a unit sphere $S^2$ and a hyperbolic surface $X$. Then the product $S^ 2 \times X$ is not flat and has zero scalar curvature. Also it is conformally flat by a paper
Simon Salamon (2009) Complex structures and conformal geometry. In: BOLLETTINO DELLA UNIONE MATEMATICA ITALIANA, vol. 9, pp. 199-224
as per editor Holonomia.
A: Responding to the OP's comments on the duplicate question, the example that was given in dimension $4$ can be easily generalized to higher dimensions:

If $(M,g_M)$ has constant sectional curvature $+1$ and $(N,g_N)$ has constant sectional curvature $-1$, then the product manifold $(M\times N,g_M\oplus g_N)$ is conformally flat. This is proved, e.g., in Besse's book "Einstein manifolds" (see Example 1.167, p. 61).
  Furthermore, this product manifold has vanishing scalar curvature if and only if $\dim M=\dim N$.

A: It seems to me that the answer is NO. Namely, if $(M,g)$ is compact, conformally flat and has zero scalar curvature then $g$ is a flat metric. Indeed, by the hypothesis there is a smooth function $f:M \to \mathbb R$ such that $e^f g$ is a flat metric on $M$. Then a finite covering of $M$ is a torus. The metric $g$ then lift to such a torus and by the Gromov-Lawson theorem (see Corollary A in http://www.ihes.fr/~gromov/PDF/8%5B26%5D.pdf) the lifted metric is flat hence the original $g$ is flat. QED
As Robert Bryant observed I made a mistake assuming that $e^f g$ is a flat metric on $M$. Indeed, conformally flat means that $e^f g$ has constant sectional curvatures so the sectional curvature can be also 1 or -1. Watching the formula for the scalar curvature of a conformal change 
https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry
it seems that the problem reduce to the question if the number $(n-2)/4(n-1)$ is an eigenvalue of the Laplacian of a compact hyperbolic manifold of dimension $n$. 
The same formula allows to rule out the flat case without using Gromov-Lawson theorem. Indeed, if $e^f g$ is flat and $g$ has zero scalar curvature then $e^{n-2}f$ is harmonic hence constant since $M$ is compact. So $g$ is flat.
