The answer is always no unless $G$ is trivial. In fact, I can generalize your statement slightly: I only need to assume that $G$ and $BG$ have merely the homotopy type of finite complexes.

I will outline the argument below for the case when $G$ is connected. (I know how to extend the argument I give below to disconnected $G$, although it is not published). My argument is purely homotopy theoretic.

In my paper, "The dualizing spectrum of a topological group" (here's a link: http://www.math.wayne.edu/~klein/quinn.pdf) I introduced the dualizing spectrum $D_G$ of a topological group. This is given by the homotopy fixed points of the left translation action of $G$ on the suspension spectrum of $G_+ = G \amalg \text{pt}$. That is, $$ D_G := \text{maps}_G(EG,\Sigma^\infty G_+) \, . $$ I then showed that if $$ F \to E \to B $$ is a fibration with $F,E,B$ homotopy finite, connected and based, then $$ D_{\Omega E } \simeq D_{\Omega B } \wedge D_{\Omega F } $$ where $\Omega E$ is a suitable topological group model for the loop space of $E$.

Let's consider the case of a connected topological group $G$ such that both $G$ and $BG$ are homotopy finite (finitely dominated is actually good enough here).

Then we have the universal bundle $$ G \to EG \to BG $$ and we get $$ S^0 \simeq D_{\Omega G} \wedge D_G $$ (here I'm using the fact that $D_{\Omega EG}$ is the sphere spectrum up to homotopy--this is not hard to check).

From the last equation, it's relatively straightforward to check that both $D_G$ and $D_{\Omega G}$ are sphere spectra up to homotopy. Furthermore, if $D_G \simeq S^{-d}$ it must be the case that $D_{\Omega G} \simeq S^d$.

But another theorem in my paper shows that whenever $G$ is a topological group with $BG$ finitely dominated, then $BG$ is a Poincaré space of dimension $d$ if and only if $D_G$ is weak equivalent to $S^{-d}$.

It follows from this that $G$ is a Poincaré space of dimension $d$ and $BG$ is a Poincar'e space of dimension $-d$.

The only possibility then is that $G$ and $BG$ are Poincaré spaces of dimension $0$. But since we are assuming that $G$ is connected, it follows that $G$ is trivial.

The answer is always no unless $G$ is trivial. In fact, I can generalize your statement slightly: I only need to assume that $G$ and $BG$ have merely the homotopy type of finite complexes.

I will outline the argument below for the case when $G$ is connected. (I know how to extend the argument I give below to disconnected $G$, although it is not published). My argument is purely homotopy theoretic.

In my paper, "The dualizing spectrum of a topological group" (here's a link: http://www.math.wayne.edu/~klein/quinn.pdf) I introduced the dualizing spectrum $D_G$ of a topological group. This is given by the homotopy fixed points of the left translation action of $G$ on the suspension spectrum of $G_+ = G \amalg \text{pt}$. That is, $$ D_G := \text{maps}_G(EG,\Sigma^\infty G_+) \, . $$ I then showed that if $$ F \to E \to B $$ is a fibration with $F,E,B$ homotopy finite, connected and based, then $$ D_{\Omega E } \simeq D_{\Omega B } \wedge D_{\Omega F } $$ where $\Omega E$ is a suitable topological group model for the loop space of $E$.

Let's consider the case of a connected topological group $G$ such that both $G$ and $BG$ are homotopy finite (finitely dominated is actually good enough here).

Then we have the universal bundle $$ G \to EG \to BG $$ and we get $$ S^0 \simeq D_{\Omega G} \wedge D_G $$ (here I'm using the fact that $D_{\Omega EG}$ is the sphere spectrum up to homotopy--this is not hard to check).

From the last equation, it's relatively straightforward to check that both $D_G$ and $D_{\Omega G}$ are sphere spectra up to homotopy. Furthermore, if $D_G \simeq S^{-d}$ it must be the case that $D_{\Omega G} \simeq S^d$.

But another theorem in my paper shows that whenever $G$ is a topological group with $BG$ finitely dominated, then $BG$ is a Poincaré space of dimension $d$ if and only if $D_G$ is weak equivalent to $S^{-d}$.

It follows from this that $G$ is a Poincaré space of dimension $d$ and $BG$ is a Poincar'e space of dimension $-d$.

The only possibility then is that $G$ and $BG$ are Poincaré spaces of dimension $0$. But since we are assuming that $G$ is connected, it follows that $G$ has the homotopy type of a point.

The answer is always no unless $G$ is trivial. In fact, I can generalize your statement slightly: I only need to assume that $G$ and $BG$ have merely the homotopy type of finite complexes.

I will outline the argument below for the case when $G$ is connected. (I know how to extend the argument I give below to disconnected $G$, although it is not published). My argument is purely homotopy theoretic.

In my paper, "The dualizing spectrum of a topological group" (here's a link: http://www.math.wayne.edu/~klein/quinn.pdf) I introduced the dualizing spectrum $D_G$ of a topological group. This is given by the homotopy fixed points of the left translation action of $G$ on the suspension spectrum of $G_+ = G \amalg \text{pt}$. That is, $$ D_G := \text{maps}_G(EG,\Sigma^\infty G_+) \, . $$ I then showed that if $$ F \to E \to B $$ is a fibration with $F,E,B$ homotopy finite, connected and based, then $$ D_{\Omega E } \simeq D_{\Omega B } \wedge D_{\Omega F } $$ where $\Omega E$ is a suitable topological group model for the loop space of $E$.

Let's consider the case of a connected topological group $G$ such that both $G$ and $BG$ are homotopy finite (finitely dominated is actually good enough here).

Then we have the universal bundle $$ G \to EG \to BG $$ and we get $$ S^0 \simeq D_{\Omega G} \wedge D_G $$ (here I'm using the fact that $D_{\Omega EG}$ is the sphere spectrum up to homotopy--this is not hard to check).

From the last equation, it's relatively straightforward to check that both $D_G$ and $D_{\Omega G}$ are sphere spectra up to homotopy. Furtheremore, if $D_G \simeq S^{-d}$ it must be the case that $D_{\Omega G} \simeq S^d$.

But another theorem in my paper shows that whenever $G$ is a topological group with $BG$ finitely dominated, then $BG$ is a Poincar'e space of dimension $d$ if and only if $D_G$ is weak equivalent to $S^{-d}$.

It follows from this that $G$ is a Poincar'e space of dimension $d$ and $BG$ is a Poincar'e space of dimension $-d$.

The only possibility then is that $G$ and $BG$ are Poincare spaces of dimension $0$. But since we are assuming that $G$ is connected, it follows that $G$ is trivial.

The answer is always no unless $G$ is trivial. In fact, I can generalize your statement slightly: I only need to assume that $G$ and $BG$ have merely the homotopy type of finite complexes.

I will outline the argument below for the case when $G$ is connected. (I know how to extend the argument I give below to disconnected $G$, although it is not published). My argument is purely homotopy theoretic.

In my paper, "The dualizing spectrum of a topological group" (here's a link: http://www.math.wayne.edu/~klein/quinn.pdf) I introduced the dualizing spectrum $D_G$ of a topological group. This is given by the homotopy fixed points of the left translation action of $G$ on the suspension spectrum of $G_+ = G \amalg \text{pt}$. That is, $$ D_G := \text{maps}_G(EG,\Sigma^\infty G_+) \, . $$ I then showed that if $$ F \to E \to B $$ is a fibration with $F,E,B$ homotopy finite, connected and based, then $$ D_{\Omega E } \simeq D_{\Omega B } \wedge D_{\Omega F } $$ where $\Omega E$ is a suitable topological group model for the loop space of $E$.

Let's consider the case of a connected topological group $G$ such that both $G$ and $BG$ are homotopy finite (finitely dominated is actually good enough here).

Then we have the universal bundle $$ G \to EG \to BG $$ and we get $$ S^0 \simeq D_{\Omega G} \wedge D_G $$ (here I'm using the fact that $D_{\Omega EG}$ is the sphere spectrum up to homotopy--this is not hard to check).

From the last equation, it's relatively straightforward to check that both $D_G$ and $D_{\Omega G}$ are sphere spectra up to homotopy. Furthermore, if $D_G \simeq S^{-d}$ it must be the case that $D_{\Omega G} \simeq S^d$.

But another theorem in my paper shows that whenever $G$ is a topological group with $BG$ finitely dominated, then $BG$ is a Poincaré space of dimension $d$ if and only if $D_G$ is weak equivalent to $S^{-d}$.

It follows from this that $G$ is a Poincaré space of dimension $d$ and $BG$ is a Poincar'e space of dimension $-d$.

The only possibility then is that $G$ and $BG$ are Poincaré spaces of dimension $0$. But since we are assuming that $G$ is connected, it follows that $G$ is trivial.