Additive category that is not abelian What is a simple example, without getting into the mess of triangulated categories, of an additive category that is not abelian?
 A: There are at least two kinds of (interesting) examples.
I. When we can get an abelian category, but have to add more (co)kernels: 1) the category of free modules over a ring; 2) the category of projective modules over a ring; 3) the category of vector bundles on a topological space (if fact, 3 is a particular case of 2).
(From 1 or 2 one gets abelian category of all modules over the ring, from 3 — abelian category of sheafs of vector spaces on X.)
II. When we already have (co)kernels ("category is pre-abelian") but not all mono-/epimorphisms are normal. As explained in another answer, an example is the category of filtered modules.
A: Another nice example is the category of finite-dimensional vector bundles over a fixed base space (with bundle maps over the identity as morphisms).
If the base space is not too simple (the interval suffices), then this category is not (pre-)Abelian because there are, in general, neither kernels nor cokernels. Intuitively speaking, the obvious candidate for the kernel of a bundle map need not form a vector bundle because its dimension need not be locally constant.
A: There's a slight modification of an example that almost works but that was deleted: take the category of Hausdorff topological abelian groups. The coimage of a morphism in this category is its image in the ordinary sense (edit: topologized via the quotient topology), but the image is the closure of the coimage (exercise) (edit: topologized via the subspace topology), so the two don't need to agree in general.
(Edit, 6/21/22: In fact the difference between the subspace and quotient topologies means the category of topological abelian groups already works, no Hausdorff assumption necessary.)
As Matt E mentions in a comment, adding topologies, like adding filtrations, is an easy way to cause this sort of thing to happen. Similarly we could take Hausdorff topological vector spaces, etc.
A: There've been lots of mildly complicated examples given, but what about the category of even-dimensional vector spaces over a field?
A: A quite explicit example coming from quantum algebra. 
Consider the ring $K_h:=\mathbb K[[h]]$ of formal power series with coefficients in  the field $\mathbb K$ (take for example $K=\mathbb R)$. Let $\mathcal C_f$ be the category of topologically free $K_h$ modules, i.e. all those $K_h$-modules that are isomorphic to modules of the form $M[[h]]$, denoting by $M$ any $K$-vector space. Morphisms $\varphi : M[[h]]\rightarrow N[[h]]$ are formal power series
$\sum_{i\geq 0} \varphi_i h^i$ with $\varphi_i: M\rightarrow N$  morphism of $K$-vector spaces.
If  $\varphi : M[[h]]\rightarrow N[[h]]$ with 
$\varphi=\sum_{i\geq 0} \varphi_i h^i$ and $\psi : N[[h]]\rightarrow Q[[h]]$ with 
$\psi=\sum_{i\geq 0} \psi_i h^i$ are morphisms in $\mathcal C_f$, then their composition is the morphism $\psi\circ\varphi$ with power series expansion
$\sum_{i+j\geq 0}(\psi_j\circ\varphi_i) h^{i+j}$.
$\mathcal C_f$ is additive with biproduct $M[[h]]\oplus_h N[[h]]:=(M\oplus N)[[h]]$;
it is not abelian because the cokernel of the inclusion (which is a morphism in $\mathcal C_f$)
$i: M\rightarrow M[[h]]$, $m\mapsto i(m)=(m,0,0,\dots)$
i.e. $coker(i)=hM[[h]]$, is not an object in $\mathcal C_f$. In fact, there exists no $\mathbb K$-vector space $N$ and isomorphism $\rho: hM[[h]] \rightarrow N[[h]]$ in $\mathcal C_f$; if such a morphism $\rho$ existed, then it would not be an isomorphism as any object in  $N[[h]]$ of the form $(n,0,0,...)$ does not belong to the 
image of $\rho$ for $n\neq 0$.
A: In infinite dimensions, all hell breaks loose. For example, neither the category of Banach spaces nor the category of Hilbert spaces, although additive, are abelian.
A: The category of finitely generated modules over a non-Noetherian ring.
The category of filtered modules over a ring is an example given in Gelfand-Manin.
