One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the von Neumann algebras they generate must have nontrivial projections (unless it's just the complex numbers, of course). A good example of this is the reduced group $C^\ast$-algebra of any free group $F_n$. If $n=1$, then $C_r^\ast(Z)\cong C(S^1)$ via the Gelfand transform, which is clearly projectionless. If $n\geq 2$, the proof is fairly complicated. See Davidson's book for a proof when $n=2$.
If $G$ is a torsion-free group, is the reduced group $C^\ast$-algebra of $G$ projectionless? This $C^\ast$-algebra always contains the group algebra $C[G]$, so a simpler question is whether $C[G]$ is projectionless if $G$ is torsion-free.
Note that torsion-free is a necessary condition as one gets a projection from summing up the elements in the cyclic group generated by a torsion element and dividing by the order of the element.
EDIT: changed typestting. still some bugs... help please?