How many cells needed to build the classifying space $BG$? Let $G$ be a finitely presented group of cohomological dimension $n$.
Apart from the unresolved ambiguity pertaining to the Eilenberg--Ganea conjecture, it is known that we can find an $n$-dimensional model of the classifying space $BG = K(G,1)$.
It is also not hard to see that $BG$ needs to have cells in every dimension $j \leq n$: otherwise $EG^{(j-1)}$ would be a simply-connected acyclic space, hence already equal to $EG$, which is impossible by the assumption on $\text{cd}(G)$.
My question now is: what is the smallest number of $j$-cells that $BG$ can have?
For $G = \mathbf{Z}^n$ the count is $\binom{n}{j}$, and in general the number of $1$- and $2$-cells are the minimal number of generators and relations of a presentation of $G$, so can be $2$ (except for $G = \mathbf Z$ where $n = 1$) and $1$, respectively.
 A: A group $G$ is of type $\mathcal{F}_n$ if it has a $K(G,1)$ with finite $n$-skeleton. Let $F_2$ be the free group on $2$ generators. Consider the kernel of the map $F_2\times F_2 \times \cdots \times F_2 \to \mathbb{Z}$, sending each generator to $1$. Then the kernel is of type $\mathcal{F}_{n-1}$ but not of type $\mathcal{F}_n$. This was proved by Stallings for $n=2$ and Bieri for $n>2$, see the discussion and links here. Taking $n\geq 3$, one gets finitely presented groups of finite cohomological dimension which do not have finite $n$-skeleton.
Nevertheless, one can ask your question for groups of cohomological dimension $n$ and type $\mathcal{F}_n$. For surface groups, and more generally 1-relator groups without torsion, this is known. A 1-relator group $G$ without torsion has presentation complex a $K(G,1)$, and cannot have a complex with fewer cells unless it is free.
For 3-manifold groups, this is a well-studied question when the $K(G,1)$ is a manifold and the cell structure is a handle structure. In this case, the number of cells is determined by the Heegaard genus of the manifold. There are known algorithms to compute this. However, the rank and the Heegaard genus can differ. So there can be a $K(G,1)$ with fewer $1$-cells than the Heegaard genus. Most likely the rank of $3$-manifolds (so the minimal number of $1$-cells in a $K(G,1)$) is computable for $3$-manifolds, but it has only been proved in special cases. I'm not sure though whether the minimality of the number of 2-cells has been addressed. The minimal number of 3-cells is $1$. An interesting possibility is that for a CW complex with a minimal number of $2$-cells, the number of $3$-cells or 1-cells might not be minimal.
In general, I'm not sure how much is known about this question, I'm just addressing the few special cases that I am familiar with to show how complex the answer can be.
