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 $C$...

thestable homotopy category, I mean the category of spectra. I hope the way I rewrote the question helps. – Don Stanley Mar 3 '10 at 15:22