When are almost complex structures tame? Let $(M^{2n},\omega)$ be a symplectic manifold. 
It is a standard result that the spaces $\mathcal{J}_{\tau}(M,\omega)$ and $\mathcal{J}(M,\omega)$ of $\omega$-tame and $\omega$-compatible almost complex structures on $M$ are nonempty and contractible. 
Suppose now that we instead fix an almost complex structure $J$ on $M$ and there exists a nondegenerate 2-form $\beta$ which tames $J$. Does there necessarily exist a symplectic form (a nondegenerate and closed 2-form) which tames $J$?
More generally, are there any sufficient conditions for the existence of a symplectic structure which tames a given almost complex structure?
 A: An almost complex structure on a manifold is not necessarily tamed by a symplectic form (except of course in real dimension 2).
For instance there exist complex manifolds, i.e. manifolds that support (integrable almost) complex structures, which do not support any symplectic structure (hence are not Kaehler). The typical example is the (complex) quotient manifold $(\mathbb{C}^4 \setminus \{0\})/\{z \sim 2z\}$ which is diffeomorphic to $S^3 \times S^1$.
If one restricts attention to manifolds $M$ which admit a symplectic form, the answer is still no. The rough reason for this fact is that almost complex structures are extremely flexible objects in real dimension greater than 2, whereas symplectic forms are not. I shall sketch how to produce a counter-example in a local chart, morally by exhibiting a locally defined almost complex structure $J$ which admits a nonconstant null-homologous $J$-curve (which, by Stoke's theorem, is incompatible with the presence of a symplectic form taming $J$).
Step 1: Given a moving frame $x \mapsto F(x) = (f_1(x), \dots, f_{2n}(x))$ on an open set $U \subset \mathbb{R}^{2n}$, we can produce an almost complex structure $J_F$ on $U$ as follows: for each each $x \in U$ and for each $i = 1, \dots, n$ set $J_F(f_{2i-1}) = f_{2i}$, $J_F(f_{2i}) = - f_{2i-1}$ and extend $J_F$ to $T_xU$ by linearity.
Step  2: Embed the 2-torus inside $\mathbb{R}^3$ in the most standard way. It is not difficult to show that there is a 3-frame $F' = (f'_1, f'_2, f'_3)$ along this torus such that the first two vector fields are tangent to the torus.
Step 3: The embedding $\mathbb{R}^3 \hookrightarrow \mathbb{R}^3 \times \{0\} \subset \mathbb{R}^3 \times \mathbb{R}^{2n-3}$ embeds also the torus, so we can complete $F'$ into a moving $2n$-frame $F$ along this (new) torus $T$. However we shall not choose any completion: we shall ask for it to be such that the resulting map $F : T \to \{\mbox{$2n$-frames}\} \cong Gl_{\mathbb{R}}(2n)$ be null-homotopic.
Step 4: By the last requirement, it is possible to extend $\left. F \right|_T$ into a moving $2n$-frame $F$ defined over the whole $\mathbb{R}^{2n}$. Consequently, we get an almost complex structure $J_F$.
Step 5: Suppose that there exists a symplectic form $\omega$ (on a ball $B \subset \mathbb{R}^{2n}$ containing $T$) which tames $J_F$, so that $\omega_x(v, J_F v) > 0$ for all $x \in B$ and nonzero $v \in T_x B$. Since the two vector fields $f'_1$ and $f'_2 = J_F(f'_1)$ of $F$ form a moving 2-frame for $T$, it follows that $\int_T \omega > 0$. But $T = \partial N$ for some embedded 3-manifold $N$ in $\mathbb{R}^{2n}$, hence $\int_T \omega = \int_N \mathrm{d}\omega = 0$. This contradiction proves that there is no symplectic form taming $J_F$.
Remark: Steps 2-3-4 are intertwined to the extend that they could be done in other orders, for example one could choose $f'_1$ and $f'_2$ tangent to $T$ in step 2 in such a way that the trivial completion in step 3 would make $F$ null-homotopic. An argument about how to do so follows similar lines to my answer to this MSE question.
I do not know whether an almost complex structure $J$ which does not admit null-homologous nonconstant $J$-curves could be tamed by a symplectic structures; there might be other, subtler obstructions to consider. In fact, I do not know any (nontrivial) sufficient or necessary condition on $J$ for it to be tamed by a symplectic structure.
