I'm reading a paper and here the authors say that a connected 4manifold with zero rational top homology has a homotopy type of 3dimensional CWstructure. I can't figure out how it can be done.

For this to be true, you need to assume that your $4$manifold $M$ is not a compact nonorientable manifold. Otherwise, you would have $H_4(M;\mathbb{Q}) = 0$ but $H_4(M;\mathbb{Z}/2) \neq 0$, so there is no hope that your manifold is homotopy equivalent to a $3$dimensional CWcomplex. Assuming this, your conditions imply that your $4$manifold is not compact (otherwise the 4th homology group would be $\mathbb{Q}$). It is a general fact that smooth noncompact $n$manifolds are homotopy equivalent to $(n−1)$dimensional CW complexes. For details, see Mohan Ramachandran's answer to my question here. 

