Sheaves without global sections
The bundles with no derived global sections (more generally the objects $F$ of the derived category $D^b(coh X)$ such that $Ext^\bullet(O_X,F) = 0$) form the left orthogonal complement to the structure sheaf $O_X$. It is denoted $O_X^\perp$. This is quite an interesting subcategory of the derived category.
For example, if $O_X$ itself has no higher cohomology (i.e. it is exceptional) then there is a semiorthogonal decomposition $D^b(coh X) =< O_X^\perp, O_X >$. Then every object can be split into components with respect to this decomposition and so many questions about $D^b(coh X)$ can be reduced to $O_X^\perp$ which is smaller. Further, if you have an object $E$ in $O_X^\perp$ which has no higher self-exts (like $O(-1)$ on $P^2$), you can continue simplifying your category --- considering a semiorthogonal decomposition $O_X^\perp = < E^\perp, E >$. For example if $X = P^2$ and $E = O(-1)$ then $E^\perp$ is generated by $O(-2)$, so there is a semiorthogonal decomposition $D^b(coh P^2) = < O_X(-2), O_X(-1), O_X >$ also known as a full exceptional collection on $P^2$. It allows a reduction of many problems about $D^b(coh P^2)$ to linear algebra.
Another interesting question is when $O_X$ is spherical (i.e. its cohomology algebra is isomorphic to the cohomology of a topological sphere). This holds for example for K3 surfaces. Then there is a so called spherical twist functor for which $O_X^\perp$ is the fixed subcategory.
Thus, as you see, the importance of the category $O_X^\perp$ depends on the properties of the sheaf $O_X$.
This response doesn't quite answer your question, but it may provide some interesting further examples.
A supernatural sheaf on $\mathbb P^n$ (really some twist of such a sheaf) has this property, though supernaturality is much stronger than you are looking for. A supernatural sheaf $\mathcal F$ is defined by two conditions:
For every $j\in \mathbb Z$, there exists at most one such $i$ such that $$ H^i(\mathbb P^n, \mathcal F(j))\ne 0. $$
The Hilbert polynomial $\chi(\mathbb P^n, \mathcal F)$ has distinct integral roots.
Hence, if $\mathcal F$ is supernatural and $e$ is a root of $\chi(\mathbb P^n,\mathcal F)$, then $\mathcal F(e)$ has the property that $H^i(\mathbb P^n, \mathcal F(e))=0$ for all $i$.
This is really the answer to a different question, but Eisenbud and Schreyer show that these sheaves can be used to generate the cohomology table of any sheaf. In addition, you could probably apply their constructions to build further examples of sheaves with the property that you are looking for. See
http://arxiv.org/abs/0712.1843 or http://arxiv.org/abs/0902.1594.
Hope this helps.
I haven't really thought about it, so I can't say whether there is good description in general. But here a class of examples, which suggests to me that such the set of such sheaves of fixed rank won't always form a bounded family. Take a smooth rational surface $X$, and choose a smooth rational curve $C\subset X$. The ideal sheaf fits into an exact sequence $$0\to \mathcal{O}_X(-C)\to \mathcal{O}_X\to \mathcal{O}_C\to 0$$ Writing out cohomology, one sees that $H^i(X,\mathcal{O}_X(-C))=0$ for all $i$.
Edit It's interesting to read other people's perspectives on this. But, anyway, for some reason I didn't finish my thought yesterday. To get an unbounded family, take $X$ to be a rational surface with infinitely many exceptional curves $C$ (which exists thanks to Nagata). Then the collection $\{\mathcal{O}_X(-C)\}$ gives an infinite set of such sheaves with distinct Chern classes.