Is a space with no nontrival vector bundles contractible? Let $X$ be a "nice" space, say having the homotopy type of a CW complex. Suppose also that $X$ is connected. Suppose that all real vector bundles on $X$ are trivial. Does it follow that $X$ is contractible?
If not, is there an example where $X$ is very nice, say a compact smooth manifold?
Here is an idea I had for the construction of a counter example. Let $G$ be a group having no nontrivial representations into any real finite-dimensional vector space. For instance, one can take $G = \varinjlim A_n$, the "alternating group on $\mathbf N$" consisting of those permutations of $\mathbf N$ fixing almost all elements pointwise and having sign $+1$. Let $X$ be a $K(G,1)$. Then I believe that $X$ has no nontrivial vector bundles, though I am not certain of my argument (the essential point of which is the equality $$\check{H^1}(X, \text{GL}_n(\mathbf R)) \overset{?}{=} \text{Hom}(G, \text{GL}_n(\mathbf R).)$$
I also thought of using the Atiyah-Hirzebruch spectral sequence to produce a counter-example, but I didn't come up with anything concrete. Anyways, I think it's an interesting question.
 A: It is not the case that $X$ must be contractible. For example, take $X = S^3$. By the clutching construction, the isomorphism classes of real rank $n$ vector bundles on $S^3$ are given by $[S^2, O(n)] = \pi_2(O(n)) = 0$ as $O(n)$ is a Lie group. More generally, every principal $G$-bundle on $S^3$ is trivial for any Lie group $G$.
A: $X = S^0$, i.e a space consisting of precisely two points (with the discrete topology).  This supports only trivial bundles and it is not contractible.  It's also a compact $0$-dimensional manifold.  
I think there's connected examples as well but I just noticed you hadn't ruled that out.  
edit: take $X$ to be the $2$-complex associated to the presentation of the binary icosahedral group:
$$ X =  <s,t | s^{-3}(st)^2, t^{-5}(st)^2> $$
This is a 2-complex such that all of its homology groups are trivial, i.e. $H_i X = 0$ for all $i \geq 1$.  Perhaps the simplest way to see this is that it's the once-punctured Poincare dodecahedral space. 
The claim is that any bundle over $X$ must be trivial. 
$\pi_1 X$ is the binary icosahedral group (the fundamental group of the Poincare homology sphere), which has trivial abelianisation. 
Given a bundle over $X$ it is classified by a map $X \to BO_k$.  There is the extension $SO_k \to O_k \to \mathbb Z_2$.  Since $\pi_1 X$ has no epimorphism to $\mathbb Z_2$, the map $X \to BO_k$ lifts to a map $X \to BSO_k$.  The first non-trivial homotopy group of $BSO_k$ is $\pi_2 BSO_k = \pi_1 SO_k$ which is cyclic. 
Now you can ask if this map is null-homotopic?  The idea is we can null-homotope $X \to BSO_k$ on the 1-skeleton since $\pi_1 BSO_k$ is trivial.  The obstruction to null-homotoping over the 2-skeleton is an element of $H^2 X$ with either $\mathbb Z$ or $\mathbb Z_2$ coefficients, but we know this group is trivial. 
So I think that does the job.  You can complete this to a rather direct "construction" of a trivialization of the bundle where one plays games, sliding obstructions around the 2-cells, playing with the above relators. 
