Let $M$ denote an arbitrary closed, connected, n-dimensional manifold for $n\geq 4$. Does there always exist a closed (not necessarily connected!) codimension two submanifold $S \subset M$ such that $\pi_1(M\setminus S)$ is a free group? If not, what if one replaces $M$ by $M\# N$ for some other $n$-manifold $N$?

A couple of thoughts on this: 

 - By playing around with a nice triangulation of $M$ one can easily construct a codimension two subcomplex of $M$ such that its complement has free fundamental group. But of course resolving the singularities seems complicated, this is why the connected sum might be helpful.

- If $n$ is big enough, I think, that the number of connected components of $S$ bounds the number of relations in $\pi_1(M)$. Because then the complement can be arranged to have a handle decompositon without two-handles and then the only two-handles of $M$ come from the $n-2$ handles in $S$.

- If $S$ has simply-connected components, then the canonical map $M \to B\pi_1(M)$ factorizes through a complex of geometric dimension two, which is impossible for many $M$.

- I think if $M$ has dimension three, then this is impossible since there are too few manifolds with free fundamental group.

- Finally if $M$ is simply-connected, then this is easy.