Trivial homology with local system Let $X$ be the classifying space of the Higman group $G$. It is well known that $G$ is an acyclic group 
$$H_{\ast}(X;\mathbb{Z})=H_{\ast}(pt;\mathbb{Z}).$$
Now, suppose that $\mathcal{M}$ is a  local system on the space $X$ such that 
$$H_{i}(X;\mathcal{M})=0, \textrm{ for all $0\leq i$}.$$ 
Does such local system $\mathcal{M}$ on $X$ exist (other then $\mathcal{M}= 0$)
 A: For $X = BG$ local systems on $X$ can be identified with $G$-modules, and homology with the derived tensor product $-\otimes^L_{\mathbb ZG}\mathbb Z$, i.e. $H_i(X;M) \cong \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. One way to see this is to take the definition $H_i(X;M):= H_i(\mathcal S_*(\widetilde X)\otimes_{\mathbb Z\pi_1(X)} M)$, where $\pi_1(X)$ acts on (singular) chains on the universal cover $\widetilde X$ via deck transformations, and replace $\mathcal S_*(\widetilde X)$ with the cellular complex $C_*(\widetilde X)$ of the CW structure of the realization of the nerve of the groupoid $G//G$; then $C_*(\widetilde X)\otimes_{\mathbb Z G} M = \dots\to \mathbb Z[G^2]\otimes M\to \mathbb Z[G]\otimes M\to M$ is the bar complex computing group homology.
Let $G$ be an arbitrary group, and let $M = \operatorname{ker}(\mathbb Z G\to \mathbb Z)$ be the reduced group algebra, so that we have a short exact sequence
$$
0\to M\to \mathbb ZG\to\mathbb Z\to 0
$$
of $\mathbb ZG$-modules. This gives rise to a long exact sequence of Tor-groups, in particular a boundary operator $\partial:\operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)\to \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. Its kernel is the image of the map $0 = \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, so it is always injective; its cokernel is the kernel of the map $\operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, which is an isomorphism for $i = 0$ and has the zero group as its codomain for $i > 0$, so it is always injective, so $\partial$ is always surjective and thus always an isomorphism.
In your example of the Higman group, we know that $H_i(X;\mathbb Z)$ is $\mathbb Z$ concentrated in degree $0$, so that $\partial$ is an isomorphism with the zero group, so that $H_i(X;M) = 0$ for all $i \ge 0$.
