Let $X$ be a connected $CW$-complex, such $\pi\_1(X)$ is torsion-free and $H\_k(X,\mathbb Z) = 0$ for all $k \geq N$ and some $N \in \mathbb N$. Then 


$(1)$ Does it follow that $X$ is homotopy-equivalent to its $N$-skeleton, i.e $X \simeq X^{(N)}$ ?


$(2)$ If $(1)$ is false, does it follow that $X \simeq X^{(k)}$ for some $k \in \mathbb N$ ? 


$(3)$ If $(2)$ is also false, at the very least, does it follow that $X \simeq Y$ for some finite-dimensional $CW$-complex $Y$ ? 


It is clear to me that the question has a negative answer when $H\_k$ is replaced by $\pi\_k$ (take a $CW$-model of $BG$ for a group $G$ containing elements of finite order). I am also very certain that this question has been asked before and I just couldn't find the corresponding thread, so I have no problem with this question being marked as duplicate. 


Edit: One could then possibly find a *finitely dominated* $CW$-complex $X$, with $\pi\_1(X)$ not torsion-free, such that its *[Wall finiteness obstruction](https://en.wikipedia.org/wiki/Wall%27s_finiteness_obstruction)* element $w(X) \in \widetilde{K\_0}(\mathbb Z[\pi\_1(X)])$ is non-trivial. Any finitely dominated $CW$-complex has finite homological-dimension (so it satisfies the assumptions), but if its finiteness obstruction doesn't vanish, it doesn't have the homotopy type of a finite complex. 


Conversely, if $X$ is finitely-dominated and $w(X) = 0$, then it has the homotopy type of a finite-dimensional complex.


In particular, this must be the case if $\widetilde{K\_0}(\mathbb Z[\pi\_1(X)]) =0$. It is conjectured that $\widetilde{K\_0}(\mathbb Z[G]) = 0$ if $G$ is torsion-free, hence the assumption in the statement of the question.


**Edit 2: Mark has given an example of a $CW$-complex $X$ with finite homological dimension, but infinite *co*homological dimension (and thus, has given a negative answer to my original questions) . It would be still interesting to know if there exist counter-examples $X$, such that for *any* $\pi\_1(X)$-module $M$, *both* the homology $H\_\*(X,M)$ *and* the cohomology $H^\*(X,M)$ have groups only in finitely many dimensions.** 



On spaces with finite homological dimension
You can take $X=BG$ where $G$ is a torsion-free, acyclic group of infinite cohomological dimension. Acyclic means $H\_k(BG;\mathbb{Z})=0$ for $k>0$, and infinite cohomological dimension implies infinite geometric dimension, so $BG$ is not homotopy equivalent to any finite dimensional CW complex $Y$. Such a group therefore gives a (rather extreme) counter-example to (1), (2) and (3).


I believe an infinite direct product of copies of [Higman's group](https://en.wikipedia.org/wiki/Higman_group) $H$ gives such a $G$. It is obviously torsion-free (since $H$ is), and is acyclic since homology commutes with direct limits and a finite product of acyclic groups is acyclic. Since $H$ is of type $FP\_\infty$ (iirc, the presentation complex is a model for $BH$), we can apply Benjamin Steinberg's answer [here](https://mathoverflow.net/q/272755) to conclude that $G$ has infinite cohomological dimension.


