Let $M$ be a smooth manifold.
Then, is any open covering $\{U_i\}$ of $M$ is contractible?, i.e. each $U_i$ is contractible?, where $\{(U_i, \phi_a)\}$ is an atlas of $M$, i.e. $M = \bigcup_i U_i$ and $\phi_a:U_a \mapsto \phi_a(U_a)$ is a homeomorphic.
Nothing (other than perhaps the particular definition you are using) is stopping you from covering your manifold with copies of $\Bbb R^n-\{0\}$. This can be done to any manifold.
For instance, if you have a covering using only copies of $\Bbb R^n$, then for each chart map $\phi:\Bbb R^n\to M$ you can split it into two charts by restricting $\phi$. One map $\Bbb R^n-\{0\}\to M$ and one map $\Bbb R^n-\{p\}\to M$ for some non-origin $p$. Doing this for each chart yields an atlas with no contractible $U_i$.
Of course, if your definition requires all charts to use $\Bbb R^n$ as domain, then that's a different story entirely.