Complex line bundles over $BG$ are classified by $H^1(BG, U(1)) \cong H^2(BG, \mathbb{Z})$. On the other hand, $1$-dimensional complex representations are classified by $\text{Hom}(G, U(1))$. There's a map from the latter to the former and it need not be surjective in general.

Explicitly, take $G = SL_2(\mathbb{R})$. This group is connected, so $BG$ is simply connected, and so the universal coefficient theorem, together with Hurewicz, gives

$$H^2(BG, \mathbb{Z}) \cong \text{Hom}(H_2(BG), \mathbb{Z}) \cong \text{Hom}(\pi_2(BG), \mathbb{Z}) \cong \text{Hom}(\pi_1(G), \mathbb{Z})$$

which is $\mathbb{Z}$ since $\pi_1(SL_2(\mathbb{R})) \cong \mathbb{Z}$. On the other hand, $G$ has no nontrivial finite-dimensional complex representations, and in particular no nontrivial $1$-dimensional complex representations.

I won't vouch for this being the simplest example though. I think I used to have simpler examples in mind.

Complex line bundles over $BG$ are classified by $H^1(BG, \mathbb{C}^{\times}) \cong H^2(BG, \mathbb{Z})$. On the other hand, $1$-dimensional complex representations are classified by $\text{Hom}(G, \mathbb{C}^{\times})$. There's a map from the latter to the former and it need not be surjective in general.

Explicitly, take $G = SL_2(\mathbb{R})$. This group is connected, so $BG$ is simply connected, and so the universal coefficient theorem, together with Hurewicz, gives

$$H^2(BG, \mathbb{Z}) \cong \text{Hom}(H_2(BG), \mathbb{Z}) \cong \text{Hom}(\pi_2(BG), \mathbb{Z}) \cong \text{Hom}(\pi_1(G), \mathbb{Z})$$

which is $\mathbb{Z}$ since $\pi_1(SL_2(\mathbb{R})) \cong \mathbb{Z}$. On the other hand, $G$ has no nontrivial $1$-dimensional complex representations.

I won't vouch for this being the simplest example though. I think I used to have simpler examples in mind.