It is known that the chain homotopy category $K(\mathcal{A})$ for an abelian category $\mathcal{A}$ need not be abelian. For example, $K(\mathrm{Ab})$ is not even abelian.
Are there any known conditions on $\mathcal{A}$ which ensure that $K(\mathcal{A})$ is in fact abelian? Browsing the standard references didn't turn up anything.
$K(\mathcal{A})$ will be abelian if and only if all short exact sequences split in $\mathcal{A}$.
More generally, in any triangulated category $\mathcal{T}$, the only epimorphisms are the split ones (and dually, the only monomorphisms are split). So if $\mathcal{T}$ were abelian, then every map in $\mathcal{T}$ would have to be of the form $\mu\varepsilon$, where $\epsilon$ is a split epimorphism and $\mu$ a split monomorphism.
To prove the claim about epimorphisms in triangulated categories, suppose $X\stackrel{\alpha}{\rightarrow}Y$ is an epimorphism, and complete to an exact triangle $$X\stackrel{\alpha}{\rightarrow}Y\stackrel{\beta}{\rightarrow}Z$$ Since $\alpha$ is an epimorphism and $\beta\alpha=0$, $\beta$ must be the zero map, and so $\alpha$ is a split epimorphism.