Let $M$ be a smooth (or complex) manifold. Let $M^{(n)}$ its $n$-th Cartesian product. Let $\Sigma_n$ be the symmetric group of dimension $n$, which acts on $M^n$ in the usual way.
Question: is it possible to give $M^{(n)} / \Sigma_n$ a smooth (or complex) structure which is, in some sense, compatible with the original structure on $M$?
If $M$ is a Riemann surface / complex surface then $M^{(n)}$ is a smooth manifold. It's a fairly standard argument. But as Willie Wong mentions, generally $M^{(n)}$ isn't a manifold unless you assume more of $M$. Interestingly enough, $(S^1)^{(3)}$ is a manifold and it's a fun exercise to figure out which one it is.
For the Riemann surface case, first consider $\mathbb C^{(n)}$. This is the space of n-tuples of points in $\mathbb C$ but with the ordering forgotten. As a space, it's homeomorphic to the space of monic complex polynomials of degree $n$ -- since monic complex polynomials have $n$ roots up to multiplicity -- the bijection is given in terms of the roots of the polynomials. So you can use a fundamental domain for the Riemann surface (or some other similar argument) to show $M^{(n)}$ is a manifold when $M$ is a Riemann surface.
FYI: I didn't just invent the above argument. It's a standard argument used in setting up Heegaard-Floer theory for 3-manifolds.