Is there an additive model of the stable homotopy category? $\DeclareMathOperator\Ho{Ho}$Is there a model category $C$ on an additive category such that its homotopy category $\Ho(C)$ is the stable homotopy category of spectra and the additive structure on $\Ho(C)$ is induced from that on $C$.
Basically I want to add and subtract maps in $C$ without going to its homotopy category.
I'm not asking for $C$ to be a derived category or anything like that. Just that it should have an additive structure.
As John Palmieri pointed out I should really say what structure I want the equivalence (between $\Ho(C)$ and the stable homotopy category) to preserve. Since I do want it to be a triangulated equivalence, Cisinski indicates why this is not possible.
 A: Let's extend the question somewhat: given a commutative ring spectrum $R$, is there an additive category $C$ such that $Ho(C)$ is the homotopy category of $R$-module spectra?
I don't know the answer to your question, but I think that if we assume a little more, then this can only happen if $R$ is a product of Eilenberg-Mac Lane spectra. Here's a sketch. Let's assume that $C$ is a simplicial model category, i.e. the hom-functors $Hom_C(X,Y)$ actually take values in topological spaces and $\pi_0(Hom_C(X,Y)) = [X,Y]_R$ if $X$ is cofibrant and $Y$ is fibrant. 
Let's assume $R$ is cofibrant-fibrant. Then $Hom_C(R,R) \simeq \Omega^\infty R$. But the left hand side is an abelian group, so $\Omega^\infty R$ is equivalent to a product of Eilenberg-Mac Lane spaces.
The sphere, by the way, is of course not a product of Eilenberg-Mac Lane spectra, and $\Omega^\infty S$ is not a product of Eilenberg-Mac Lane spaces.
A: In the category $Ho(C)$, say that every map is a fibration and a cofibration, and define the weak equivalences to be the isomorphisms. This makes $Ho(C)$ into a model category; its homotopy category is $Ho(C)$, and it is certainly additive.
It's not what you want, though; can you make your question more precise?
A: The answer is: no there isn't such a thing. Here is a rough argument (a full proof would deserve a little more care).
Using the main result of
S. Schwede, The stable homotopy category is rigid, Annals of Mathematics 166 (2007), 837-863
your question is equivalent to the following: does there exist a model category $C$, which is additive, and such that $C$ is Quillen equivalent to the usual model category of spectra?
In particular, we might ask: does there exist an additive category $C$, endowed with a Quillen stable model category structure, such that the corresponding stable $(\infty,1)$-category is equivalent to the stable $(\infty,1)$-category of spectra?
Replacing $C$ by its full subcategory of cofibrant objects, your question might be reformulated as: does there exist a category of cofibrant objects $C$ (in the sense of
Ken Brown), with small sums (and such that weak equivalences are closed under small sums), and such that the corresponding $(\infty,1)$-category (obtained by inverting weak equivalence of $C$ in the sense of $(\infty,1)$-categories) is equivalent to the stable $(\infty,1)$-category of spectra? If the answer is no, then there will be no additive model category $C$ such that $Ho(C)$ is (equivalent to) the category of spectra (as a triangulated category).
So, assume there is an additive category of cofibrant objects $C$, with small sums, such that $Ho(C)$ is (equivalent to) the category $S$ of spectra (as a triangulated category). Let $C_f$ be the full subcategory of $C$ spanned by the objects which correspond to finite spectra in $S$. Then $Ho(C_f)\simeq S_f$, where, by abuse of notations, $Ho(C_f)$ is the $(\infty,1)$-category obtained from $C_f$ by inverting weak equivalences, while $S_f$ stands for the stable $(\infty,1)$-category of finite spectra (essentially the Spanier-Whitehead category of finite CW-complexes). Given any (essentially) small additive category $A$ denote by $K(A)$ the "derived $(\infty,1)$-category of $A$" (that is the $(\infty,1)$-category obtained from the category of bounded complexes of $A$, by inverting the chain homotopy equivalences). Then, the canonical functor $A\to K(A)$ (which sends an object $X$ to itself, seen as a complex concentrated in degree $0$), has the following universal property: given a stable $(\infty,1)$-category $T$, any functor $A\to T$ which sends split short exact sequences of $A$ to distinguished triangles (aka homotopy cofiber sequences) in $T$ extends uniquely into a finite colimit preserving functor $K(A)\to T$. In particular, the functor $C_f\to Ho(C_f)\simeq S_f$ extends uniquely to a finite colimit preserving functor $F:K(C_f)\to S_f$. Let $Ker(F)$ be the full $(\infty,1)$-subcategory of $K(C_f)$ spanned by objects which are sent to zero in $S_f$. Then the induced functor
$$K(C_f)/Ker(F)\to S_f$$
is an equivalence of (stable) $(\infty,1)$-categories (to see this, you may use the universal property of $S_f$: given a stable $(\infty,1)$-category $T$, a finite colimit preserving functor $S_f\to T$ is the same as an object of $T$; see Corollary 10.16 in DAG I). This implies that, for any object $X$ of $S_f$, if $X/n$ denotes the cone of the map $n:X\to X$ (multiplication by an integer $n$), then $n.X/n\simeq 0$ (see Proposition 1 in Schwede's paper Algebraic versus topological triangulated categories). But such a property is known to fail whenever $X$ is a finite spectrum for $n=2$ (see Proposition 2 in loc. cit.). Hence there isn't such a $C$...
