A result of Kawamata (Kawamata, Yujiro, Characterization of abelian varieties. Compositio Mathematica, 43 no. 2 (1981), p. 253-276) implies that, under your assumptions, $X$ is birational to an abelian variety (in fact you just need the Kodaira dimension of $X$ to be zero and the irregularity to be equal to the dimension of $X$).
Once you know that $X$ is birational to an abelian variety $A$, a Lemma of Deligne implies that if the canonical divisor on $X$ is trivial, then $X$ is in fact an abelian variety. This is not a particularly deep result. First, the rational map $f \colon X \to A$ is actually a morphism (essentially because $A$ cannot contain rational curves). Second, the morphism $f$ induces a morphism $df$ between the cotangent bundles. The determinant of $df$ is a morphism between the canonical bundles of $A$ and $X$, that are both trivial by assumption. Thus the determinant is either identically zero, or it is an isomorphism. Since the morphism $f$ is generically etale, the determinant is not identically zero. But then it is an isomorphism, so that the morphism $f$ is always etale, and we conclude that $X$ is an abelian variety.
EDIT: Ah, as Pete remarked below, I did not answer the question! The answer is "Yes"! Even under weaker assumption: namely it suffices to know that the canonical bundle is trivial and that $\dim (X) = h^1(X,\mathcal{O}_X)$.