This isn't a complete answer, but I believe it is possible by the Kan-Thurston Theorem, which says that every path connected space has the (co)homology of a $K(G,1)$. So you should be able to You can build a space with prescribed nontrivial cohomology (with $V_i$-coefficients), and then that must be isomorphic to the cohomology of some $BG_i$-space (hence group $G_i)$.
The reason I am not going to say this is a complete answer, is because due to the construction of the theorem, there might be a small problem getting the desired coefficients. The paper is entitled Every Connected Space has the Homology of a $K(\pi,1)$, by Kan and Thurston.
This isn't a complete answer, but I believe it is possible by the Kan-Thurston Theorem, which says that every path connected space has the (co)homology of a $K(G,1)$. So you should be able to build a space with prescribed nontrivial cohomology (with $V_i$-coefficients), and then that must be isomorphic to the cohomology of some $BG_i$-space (hence group $G_i)$.
The reason I am not going to say this is a complete answer, is because due to the construction of the theorem, there might be a small problem getting the desired coefficients. The paper is entitled Every Connected Space has the Homology of a $K(\pi,1)$, by Kan and Thurston.