Non-Kahler "Calabi-Yau"? Are there examples of non-Kahler complex manifolds with holomorphically trivial canonical bundle?
 A: Any non-trivial principal elliptic bundle $\pi:X \to B$ over a Calabi-Yau basis is non-Kaehler but has trivial canonical bundle (because $\mathcal{K}_X \simeq \pi^*\mathcal{K}_B$).
A: There is a reasonably extensive literature on non-Kahler Calabi-Yau threefolds. They are of interest in string theory; see for example http://xxx.lanl.gov/abs/hep-th/0301161, as well as http://xxx.lanl.gov/abs/0809.4748 for an analogue of Calabi's conjecture in this context. 
A: One can ask for non-Kähler compact manifolds with holomorphically trivial tangent bundle, and still get many examples. By a result of Wang ( Proc. AMS 5 ), these are quotients of a complex Lie group $G$ by a discrete subgroup $\Gamma$.
If the quotient is compact Kähler then $G$ must be abelian. Indeed,  every vector subspace of the Lie algebra of $G$ gives rise to a holomorphic differential form on $G/\Gamma$.  If $G$ is not abelian then one can choose a subspace not closed under the Lie bracket. The corresponding differential form is clearly not closed, what cannot happen in compact Kähler manifolds.
For a thorough study of examples of this kind see this book by Winkelmann.
A: More examples are given by nilmanifolds.
Barberis, Dotti and Verbitsky proved in Theorem 2.7 in http://www.ams.org/mathscinet-getitem?mr=2496748 that nilmanifolds endowed with invariant complex structures have trivial canonical bundle. See also Cavalcanti and Gualtieri's Theorem 3.1 in http://www.ams.org/mathscinet-getitem?mr=2131642
On the other hand, non-tori nilmanifolds never admit a Kaehler structure, because of Benson and Gordon, http://www.ams.org/mathscinet-getitem?mr=976592 , or Hasegawa, http://www.ams.org/mathscinet-getitem?mr=946638 , or ...
A: Yes, you might look at the following paper by J. Fine and D. Panov: http://arxiv.org/abs/0905.3237
A: This is covered in Andrei Halanay's answer, but it's
worth mentioning the simplest examples, which are
primary Kodaira surfaces.  For the simplest of these:
Take C^2 and quotient by the group generated by these a_k:
a_1 :  z -> z + 1
a_2 :  z -> z + i
a_3 :  w -> w + z + 1
a_4 :  w -> w - iz + i
(I think this is it.)
The quotient group is nonabelian.
Here z is the fiber and w the base.
