Since nowadays lots of research activities are happening to prove many results for compact Kähler manifolds which are known for projective varieties, I was wondering are there plenty of nonprojective Kähler manifolds? If yes, where can I find some explicit examples? I am aware of the theorem that a generic complex torus $\mathbb{C}^g/\Lambda$ is nonprojective.

4$\begingroup$ The product of a complex manifold with a nonprojective manifold will not be projective. $\endgroup$ – Michael Albanese Dec 13 '16 at 18:31

6$\begingroup$ A "very generic" K3 surface is nonprojective $\endgroup$ – Francesco Polizzi Dec 13 '16 at 18:40

$\begingroup$ @MichaelAlbanese  How to prove that result? $\endgroup$ – Bingo Dec 14 '16 at 4:17

4$\begingroup$ @Bingo $X$ is a closed submanifold of $X \times Y$. If $X \times Y$ is a closed submanifold of projective space, then $X$ is a closed submanifold of projective space. $\endgroup$ – Will Sawin Dec 14 '16 at 15:04
See Claire Voisin's amazing results on the subject, or the published version:
On the homotopy types of compact kaehler and complex projective manifolds, Inventiones Math. 157 2 (2004), 329  343.
(ArXiv) Abstract: We show that in every dimension greater than or equal to 4, there exist compact Kaehler manifolds which do not have the homotopy type of projective complex manifolds. Thus they a fortiori are not deformation equivalent to a projective manifold, which solves negatively Kodaira's problem. We give both non simply connected (of dimension at least 4) and simply connected (of dimension at least 6) such examples.

2$\begingroup$ An outline of Voisin's example: take a nonalgebraic torus $T$ with suitable endomorphism $\phi$ (all eigenvalues simple and not real), then consider the following four copies of $T$ in $T\times T$: the two axes, the diagonal and the graph of $\phi$. Blow up the (finitely many, transverse) intersection points, then blow up the proper transforms of the four $T$'s. $\endgroup$ – Neil Strickland Dec 14 '16 at 17:33
I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't find any in complex dimension one, so let's look in dimension two. By classification of surfaces, the non algebraic surfaces would have Kodaira dimensions $\kappa=0$ or $1$. The $\kappa=0$ cases are, I believe, necessarily tori or K3's (there are some other examples, but these are either algebraic or non Kähler). As already noted by you and Francesco, the algebraic surfaces form a proper subset in moduli for these cases.
The $\kappa=1$ surfaces are elliptic surfaces, and I expect that there should be plenty of non algebraic examples, although I don't have an example on hand.
Added later For my own reasons, I thought about this longer than I normally would. Here's an explicit example. Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma=\pi_1(C)$, and $\tilde C$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h:\Gamma\to E$. Define an action of $\Gamma$ on $\tilde C\times E$ by $\gamma(x,y)= (\gamma x, y+h(\gamma))$, and let $S$ be the quotient.
$S$ is Kähler. If $h$ has infinite image, then $S$ is not algebraic.
Proof. $S$ is Kähler because $\tilde C\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f:S\to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.
Additional remarks: This has $\kappa=1$ when $g>1$. When $g=1$, one can see, with a bit of thought, that $S$ is torus which contains an elliptic curve but it is not isogenous to a product (i.e. Poincar\'e reducibility fails for tori).

$\begingroup$ Addendum for my own edification: since $E$ is commutative, morphisms $h\colon \Gamma\to E$ factor through the abelianization $\Gamma^{\rm ab}$ of $\Gamma$. The presentation of $\Gamma$ or the Hurewicz theorem show that the group $\Gamma^{\rm ab}$ is isomorphic to $\mathbf Z^{2g}$, hence is free abelian, and non trivial. Consequently, there are plenty of morphisms $h\colon\Gamma \to E$ with infinite image. $\endgroup$ – ACL Dec 14 '16 at 17:58

$\begingroup$ I don't see why the algebraic dimension of $S$ is one. As a matter of fact, i think it can be equal to two in some examples. There are foliations on the product $C\times E$ transverse to $\pi$ and without algebraic leaves. This leads to a presentation of the product as a suspension with infinite $h$. $\endgroup$ – Jorge Vitório Pereira Dec 14 '16 at 23:38

$\begingroup$ See mathoverflow.net/questions/231144/… $\endgroup$ – Jorge Vitório Pereira Dec 15 '16 at 12:54

$\begingroup$ OK, I'll take a look when I get a chance. $\endgroup$ – Donu Arapura Dec 15 '16 at 13:22
Expanding a bit on Francesco Polizzi's remark, complex K3 surfaces are Kähler [1], the moduli space $M$ of complex K3 surfaces is an irreducible 20dimensional complex variety, and within $M$, the set of K3 surfaces that are algebraic varieties is a countable union of disjoint subvarieties $F_g$ for integers $g\ge2$. Here $F_g$ has dimension 19, and the K3 surfaces in $F_g$ are those that have a primitive line bundle $\mathcal L$ satisfying $c_1(\mathcal L)^2=2g2$.
So this is the analogue for K3 surfaces of the example you know for complex tori, i.e., in the moduli space of complex tori of a given dimension, the projective ones form a countable union of proper subvarieties.
[1] Siu, Y. T. (1983), Every K3 surface is Kähler, Inventiones Mathematicae 73 (1): 139–150