Let $C_*$ be a chain complex such that every $C_i$ is a torsion-free finitely generated abelian group, with $C_i=0$ for every $i<0$ and every $i>N$ for some sufficiently large integer $N$. If every homology group $H_i(C_*)$ is a torsion group, then $$\bigoplus_{i \text{ even}}C_i\cong\bigoplus_{i \text{ odd}}C_i.$$
I honestly have no clue of how to begin proving this. Any help would be appreciated
Since the $C_i$ are free, only the ranks matter, and we may tensor with $\mathbb{Q}$. Then the complex becomes exact because the homology was assumed to be torsion. Now use that the Euler characteristic of an exact complex vanishes.