Global sections of the anticanonical bundle Let $X$ be a smooth projective variety (over $\mathbb{C}$) with the canonical line bundle $K_X$. Also assume that $X$ has no global holomorphic top forms i.e. $H^0(X, K_X) = 0$.
Is it true that the anti-canonical line bundle $K_X^{-1}$ always has global holomorphic sections? That is, $H^0(X, K_X^{-1}) \neq \{0\}$? Does such a global section correspond to a meromorphic top form (giving a divisor in the divisor class of $K_X$)?
 A: A preliminary comment: you ask 

Does such a global section correspond to a meromorphic top form (giving a divisor in the divisor class of $K_X$?

No, a section of $K_X^{-1}$ would give a divisor in the class $-|K_X|$. In general, for any line bundle $L$ with associated divisor class $|L|$, a nonzero global section of $L$ gives a divisor in $|L|$. More pragmatically, why would a section of the dual of $K_X$ have anything to do with meromorphic sections of $K_X$ itself?
Now to your (edited) question: here the answer is again no. Let's give some counterexamples.
Example 1: Let $X$ be an Enriques surface. This is a particular kind of minimal surface with the property that $K_X \ncong O_X$ but $K_X^2 \cong O_X$. These two facts already say that $K_X$ cannot have any nonzero global sections, because the square of such a section would give a nonconstant section of $O_X$. Moreover, $K_X^2 \cong O_X$ means that $K_X^{-1} \cong K_X$, so $H^0(X,K_X^{-1}) = 0$ too.
Example 2: If that example seems a little tricky, here's one to convince you that this behaviour is not something special, but is more like the general case. Let $X$ be the blowup of $\mathbf P^2$ in $r \geq 10$ points which do not lie on a cubic. Certainly $K_X$ has no sections, since $\operatorname{dim} H^0(X,K_X)$ is a birational invariant, and on $\mathbf P^2$ there are no such sections. What about $K_X^{-1}$? Well, the associated divisor class is $-|K_X|=3H-E_1-\cdots-E_r$, where $H$ is the class of a line on $\mathbf P^2$ (pulled back to $X$) and $E_i$ are the classes of the exceptional divisors. An effective divisor in this class would have to be the proper transform on $X$ of a cubic curve on $\mathbf P^2$ passing through the $r$ points, but by assumption no such curve exists.
Closing remark: Here is a general remark about what one should expect in questions like this. Inside the real vector space $N^1(X) := \left(\operatorname{Pic(X)}/\equiv \right) \otimes \mathbf R$ of numerical equivalence classes of line bundles, the points represented by line bundles with nonzero global sections form a convex cone called the effective cone $\operatorname{Eff}(X)$ of $X$. The criteria of Kleiman and Nakai–Moishezon imply that the closure $\overline{\operatorname{Eff}(X)}$ is a strictly convex cone, meaning that it contains no 1-dimensional subspaces. That means that, if the Picard number of $X$ is at least 2, then the union $\overline{\operatorname{Eff}(X)} \cup - \overline{\operatorname{Eff}(X)}$ of this cone with its negative cannot fill up all of $N^1(X)$. So there will be lots of line bundles $L$ such that neither $L$ nor $L^{-1}$ have global sections. Example 2 above shows that, in particular, this can happen for $L=K_X$. (Example 1 is slightly different in nature — here $K_X$ maps to $0 \in N^1(X)$ because it is torsion.)
A: No.
 For any compact Riemann  surface $X$ of genus $g\geq2$ the anticanonical bundle is a line bundle of degree $2-2g\lt 0$ and has thus no non-zero holomorphic global section: $\Gamma (X,K_X^{-1})=0 $  
Edit
This is an answer to the original question, which did not suppose $\Gamma (X,K_X)=0 $
