If $X$ is a smooth and proper stack, does it admit a smooth and proper atlas? Fix a ground scheme $S$ (a field say).
By atlas for an algebraic stack I mean a smooth and surjective morphism $Y \to X$ from a scheme (or algebraic space or affine scheme) $Y$.
If the stack $X$ is smooth then all atlases are smooth schemes over $S$.
What about when $X$ is smooth and proper? Can I find a proper atlas?
If in general the answer is no, what about for Deligne-Mumford stacks?
 A: The answer is "no".  Consider a smooth, proper $1$-dimensional Deligne-Mumford stack over $\mathbb{C}$ which has coarse moduli space $\mathbb{P}^1$ and which has a single "stacky" point (with any nontrivial stabilizer group at that point).  
Edit.  A more precise definition of the stack is as follows.  Let $m$ be an integer, $m>1$.  Let $A$ be $\mathbb{A}^2 \setminus \{(0,0)\}$ with coordinates $x$ and $y$.  Let $\mathbb{G}_m$ act on $A$ by $t\ast (x,y) = (tx,t^m y)$.  The stack $Q_m$ is the quotient stack $[A/\mathbb{G}_m]$.  The quotient morphism $A\to Q_m$ is a smooth atlas, implying that $Q_m$ is smooth (since $A$ is smooth).  In fact, this morphism is naturally a $\mathbb{G}_m$-torsor; denote the associated invertible sheaf on $Q_m$ by $\mathcal{L}$.  It is not hard to see that this is an $m$-torsion invertible sheaf; the $\text{m}^{\text{th}}$ tensor power of $A$ as a $\mathbb{G}_m$-torsor over $Q_m$ is the quotient stack $[(\mathbb{G}_m\times A)/\mathbb{G}_m]$ where the $\mathbb{G}_m$-action is $t\cdot (u,(x,y)) = (t^mu, (tx,t^my))$.  This admits a "trivializing" morphism to $\mathbb{G}_m$, $(u,(x,y)) \mapsto u/x^m = u/y$, each defined on the appropriate open $\mathbb{G}_m\times D(X)$ or $\mathbb{G}_m\times D(y)$.  The only nontrivial stabilizer group is $\mathbf{\mu}_m$ acting on the $\mathbb{G}_m$-orbit $\{0\}\times \mathbb{G}_m$ inside $A$.  So $Q_m$ is a Deligne-Mumford stack.  Finally, $Q_m$ is finite over its coarse moduli space $\mathbb{P}^1$, where the $\mathbb{G}_m$-invariant morphism $A\to \mathbb{P}^1$ is just $(x,y) \mapsto [x^m,y]$.  
