Are compact, complex, affinely flat manifolds geodesically complete? Let $M$ be a real, even dimensional, compact manifold endowed with a symplectic form  $\omega$ and a flat, torsionless connection $\nabla$ compatible with $\omega$, that is $$\nabla \omega=0.$$
Under this assumptions, taking into account a simplified version of the  Markus conjecture, it should follow that $M$ is geodesically complete.
My question is: Under what additional geometric assumptions is the result known to be true?
One concrete example is: If in addition we assume the existence of a parallel almost complex structure $J$ on $M,$ then the result is true for $2$ dimensional and $4$ dimensional manifolds as observed by Misha.
What if instead of the the parallel almost complex structure we assume that there is a nonzero parallel vector field $V$ on $M?$ Does it follow that the manifold is geodesically complete?
 A: Let me give a standard example of a closed incomplete manifold with flat affine structure, whose 2-dimensional version is essentially the example from the comment of Misha. 
Consider $R^n\setminus 0$ and the action of the group $(Z, +)$ generated by the homothety
$$
p\mapsto 2 p.$$
The action preserves the flat affine connection, and also (if the dimension is even and the standard complex structure exists) the standard complex structure.  
The quotient $R^n\setminus 0/Z$ is homeomorphic to $S^{n-1}\times S^1$ and is compact. Since the action of $Z$ preserves the flat affine and complex structure they induce a flat affine and complex structure on the quotient. Since we took out the point $0$ from $R^n$, the universal cover is not complete so  the affine strcuture is not complete
A: Now that you added the volume requirement, you get close to the Marcus conjecture. In complex dimension 1 you then get a locally Euclidean structure and the answer is clearly positive. In complex dimension 2 it is again positive since linear holonomy then has rank 1 and in this situation Marcus conjecture was proven by Yves Carriere in1989. 
A: This is an answer to the last question of the revised version of your question which is 
"What if instead of the the parallel almost complex structure we assume that there is a nonzero parallel vector field V on M? Does it follow that the manifold is geodesically complete?"
The answer is the same as to the   question without the assumption. Indeed, having an incomplete manifold with $\omega$ parallel with respect to a flat torsionsfree connection, we can directly multiply it by $R^2$ with the standard connection and the standard symplectic structure.  The resulting  manifold remains   incomplete but now it has even two parallel vector fields.  
