All simple modules are projective $\Rightarrow$ semisimple Let $A$ be a finite dimensional algebra over a field $K$. It is clear that if $A$ is semisimple, then every simple module is projective. Does the converse hold ?
It seems false, but I can't find a counterexample. A non-semisimple algebra with this property must have a non-projective indecomposable module, but that's as far as I could go. 
 A: $A$ is a module of finite length over itself. Prove, by induction on the length, that a finitely generated module over $A$ is semisimple. In particular $A$ is semisimple.
A: Every finitely generated $A$-module has finite length, and therefore it is  projective. 
Now use a result of Auslander which says the following: the left global dimension of $A$ equals the supremum of projective dimensions of cyclic $A$-modules. This implies that the global dimension of $A$ is zero, hence every $A$-module is projective, so $A$ is semisimple.
A: Note that, if every projective module is simple. Then, in particular, $_AA$ is simple. Thus, $A$ is simple ring. (I am assuming that $A$ is finite dimensional $K$-algebra with identity)
However, if every indecomposable projective left $A$-module is simple, then by using Krull-Schmidt theorem($_AA$ is a direct sum of indecomposable projective modules and these indecomposable summands are uniquely determined up to isomorphism and permutation), $A$ is a direct sum of indecomposable projective left $A$- modules. Hence, direct sum of simple modules. Thus, $A$ is semisimple.
Following the discussion in the comments and considering the title of your question. If every simple module is projective, then by Krull Schmidt theorem and the fact that indecomposable projectives have simple head. We obtain that  the simples are isomorphic to their projective cover. Thus, $A$ is a direct sum of simples.    
