Is there a manifold $M$ such that for every $x\in M$, $M-\{x\}$ is not parallelizable but there is a finite set $S\subset M$, with $\# S>1$, such that $M-S$ is parallelizable?



A question on parallelizability
No: any finite set $S \subset M$ can be contained in the interior of an embedded closed disc in $M$, and cutting this out gives a manifold diffeomorphic to $M - \{x\}$. So if $M - S$ were parallelisable, $M-\{x\}$ would be too.


