If I have a CW-complex, is it possible to find a homotopically equivalent one that will have zero boundary operators?
It shouldn't be always possible to find such a triangulation for the initial space (consider two vertices connected with three edges), but I can neither prove the statement above nor find a counterexample.
The following doesn't work because the final step of the argument is not valid.
I think the answer is that you can do this if and only if the space has the homotopy type of a wedge of (possibly infinitely many, in different dimensions) spheres. I think the following argument works.
Recall that the differentials $d_n\colon \mathcal C_n\to \mathcal C_{n-1}$ are given by $d_n(e_\alpha^n)=\sum d_{\alpha\beta} e_\beta^{n-1}$ where the $e_\alpha^n$ and $e_\beta^{n-1}$ are the $n$- and $n-1$ cells generating (as free groups) $\mathcal C_n$ and $\mathcal C_{n-1}$.
Hence, the differentials are zero if and only if the $d_{\alpha\beta}$ are zero, but $d_{\alpha\beta}$ is the degree of the inclusion of the boundary of $e_\alpha^n$ into the $n-1$-skeleton $X_{n-1}$ followed by contracting every cell other than $e_\beta^{n-1}$. This is in total a map $S^{n-1}\to S^{n-1}$, so the degree is just how top homology gets multiplied.
Since maps of degree zero are null-homotopic, you should be able to glue all the null-homotopies together on the CW complex (yay, induction and compactness of spheres) to get that the attaching map itself is null-homotopic (nope, this is blatantly false).