Assume we have a closed symplectic manifold $M$ which is the total space of a smooth fibration by half-dimensional tori. Can we infer that $M$ is the total space of a smooth fibration by Lagrangian tori?

This doesn't need to hold. For example, if one takes a $(T^4,\omega)$ with a constant symplectic structure $\omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a homologically non-trivial $T^2\subset T^4$ such that $\int_{\omega} T^2=0$ which is impossible for general $\omega$.

One can also give counter-examples when a half-dimensional torus bundle exists on the symplectic manifold but no Lagrangian torus bundle exists for any symplectic structure on the manifold. To construct such an example consider $M^4$ that fibers over $B=T^2$ with a fiber $F=T^2$ such that the action of $\mathbb Z^2=\pi_1(B)$ on $H_1(F)=\mathbb Z^2$ contains a hyperbolic element. Such a manifold is symplectic by Thurston, but it is not a total space of a Lagrangian torus fibration by the classification of such fibrations in dimension $4$.

Just to add, that the classification of Lagrangian $T^2$ fibrations was done in https://ac.els-cdn.com/S0926224596000241/1-s2.0-S0926224596000241-main.pdf?_tid=824690ea-affb-4cd2-98a6-6d7ed7f08d9f&acdnat=1544694026_3c6783fa389437767cd9c8246616d4dd

And there is as well a very nice paper classifying Lagrangian $T^2$ fibrations over the Klein bottle. As you can see from Theorem 2.1 stated in the paper, the representation of $\pi_1(B)$ in $H^1(F)=\mathbb R^2$ (for Lagragian $T^2$ fibrations over $T^2$) is unipotent. https://arxiv.org/pdf/0909.2982.pdf