TITLE: Is paralellizability a topological invariant (Invariant under homemorphism)
QUESTION [11 upvotes]: This  MO  post  is  a  motivation to  ask:

Is paralellizability  a  topological invariant (Invariant under homeomorphism)?

REPLY [14 votes]: No. You can cook up counterexamples in high dimensions via surgery theory. The relevant paper here is Milnor, "Microbundles and differentiable structures", available here. The theorem you want is Corollary 6.4; as a corollary of this, Milnor gives an example of a (noncompact) smooth parallelizable manifold that carries a different, non-parallelizable differentiable structure.
You should be able to find closed examples using this corollary, but I don't know any off the top of my head.