To expand my comment: No it is not possible.
First suppose that $G$ is connected and that both $G$ and $BG$ have the homotopy types of finite complexes. If $G$ is not contractible, then let $k>0$ be minimal such that $\pi_k(G)$ is nontrivial. We have $H_k(G)=\pi_k(G)=\pi_{k+1}(BG)=H_{k+1}(BG)$. Choose a prime $p$ such that the latter finitely generated abelian group is nontrivial mod $p$. From now on let homology be with mod $p$ coefficients. Let $d$ be maximal such that $H_d(BG)$ is nontrivial. Let $e$ be maximal such that $H_e(G)$ is nontrivial. The spectral sequence of the fibration has a nontrivial group at $E^2_{d,e}$ that will not go away. Contradiction. So $G$ connected implies $G$ contractible.
If $G$ is not connected then $\pi_0G$ is finite. Let $G_0$ be the component of the identity in $G$. Then $BG_0$, the universal cover of $BG$, is also a finite complex, so by the previous paragraph $G_0$ is contractible. That is, $G$ is (up to homotopy) discrete and $BG$ is the classifying space of a finite group. For any cyclic subgroup $C$ of the finite group $G$, $BC$ is a finite cover of $BG$ and therefore finite. The cohomology of $BC$ is such that it cannot be equivalent to a finite complex unless $C$ is trivial. So $G$ is trivial (contractible).