Let $M$ be a Riemannian manifold. I'm wondering if there are nice conditions/obstructions or (non-Kahler) examples of the existence of a map $J \in \operatorname{End} (TM \otimes \mathbb C)$ such that


1. $J^2 = -Id$.
2. $g(JX,JY) = g(X,Y)$ for all $X,Y \in TM \otimes\mathbb C$.
3. $\nabla J = 0$ where $\nabla$ is the Levi-Civita connection.


Of course, if $J$ preserves $TM$ then $M$ is Kahler which is a pretty strong condition. But, for example, if $J$ does not preserve $TM$ then the 2-form $\omega(\cdot,\cdot) = g(J\cdot, \cdot)$ is closed and non-degenerate but could now be zero in de Rham cohomology since it's not necessarily real.



Parallel orthogonal complex structures on complexified tangent bundle.
In one sense, there are only Kähler examples, but, in another sense, there are non-Kähler examples of such $J$. Here is what I mean:


Suppose that one has the data $(M,g,J)$ as defined in the question and that they satisfy the Conditions $1$ and $2$ (we'll get to $3$ below). Let $TM\otimes\mathbb{C} = W\_+\oplus W\_-$, where $W\_+\subset TM\otimes\mathbb{C}$ is the $(+i)$-eigenspace of $J$ and $W\_-\subset TM\otimes\mathbb{C}$ is the $(-i)$-eigenspace of $J$. (This direct sum decomposition exists because of Condition $1$.)


Now, by Condition $2$, $W\_+$ and $W\_-$ must be $g$-isotropic subbundles of $TM\otimes\mathbb{C}$. Since they have trivial intersection and sum to $TM\otimes\mathbb{C}$ (on which $g$ is nondegenerate), it follows that they must each be of half the dimension of $TM\otimes\mathbb{C}$, so, in particular, $M$ must have even dimension. Moreover, each of these spaces has trivial intersection with $TM\otimes 1$ and $TM\otimes i$ (since these latter spaces have no nonzero $g$-null vectors), so it follows that there are maps $K\_\pm:TM\to TM$ such that 
$$
W\_\pm = \{\ v - i\ K\_\pm v\ \bigl|\ v\in TM\ \}.
$$
Now, the condition that $W\_\pm$ be $g$-null is equivalent to the condition that each of $K\_+$ and $K\_-$ be a $g$-orthogonal almost complex structure on $M$. The condition that these two spaces be disjoint is just that $K\_+-K\_-:TM\to TM$ have no kernel. Conversely, if $K\_\pm:TM\to TM$ are two $g$-orthogonal almost complex structures on $M$ such that $K\_+-K\_-$ has no kernel, then defining the complex subbundles $W\_\pm\subset TM\otimes\mathbb{C}$ as above and defining $J:TM\otimes\mathbb{C}\to TM\otimes \mathbb{C}$ to be the complex linear map that has $W\_+$ as the $(+i)$-eigenspace of $J$ and $W\_-$ as the $(-i)$-eigenspace of $J$ will give a triple $(M,g,J)$ that satisfies Conditions $1$ and $2$. Note that the $J$ that is so-defined comes from the almost complex structure $K\_+$ in the usual way if and only if $K\_- = - K\_+$. Note that, in general, there could be many pairs $(K\_+,K\_-)$ that satisfy these conditions without satisfying $K\_-=-K\_+$.


Now, what about Condition 3 (i.e., $\nabla J=0$)? This is equivalent to $\nabla K\_\pm = 0$, so one can get examples $(M, g, J)$ that satisfy the three Conditions without being derived from a standard Kähler structure only if $g$ has a pair of parallel complex structures $(K\_+,K\_-)$ that satisfy the above conditions but don't satisfy $K\_-=-K\_+$. This does happen; for example, if $g$ has holonomy $\mathrm{Sp}(n)\subset\mathrm{SO}(4n)$, then there is a $2$-sphere of $g$-parallel complex structures on $M$, and choosing any pair $(K\_+,K\_-)$ from that $2$-sphere that is not antipodal will allow one to generate a $(M,g,J)$ that does not come from a Kähler structure in the usual sense. However, note that the underlying Riemannian manifold *does* admit a Kähler structure, in fact, a $2$-sphere of them. The family of $(M,g,J)$ that one gets for such a manifold has dimension $4$ and contains the 'usual' Kähler structures as an embedded $2$-sphere. There can be a higher dimensional family of solution pairs if the holonomy of $g$ is a proper subset of $\mathrm{Sp}(n)$, but, in such cases, the metric will be a product (locally), and the problem reduces to working on the irreducible factors.


