TITLE: Must a manifold covered by $ S^n $ admit a metric of constant positive sectional curvature?
QUESTION [10 upvotes]: Suppose that the smooth manifold $ M $ has the n-sphere for its universal cover (in the topological sense). Does there exist a Riemannian metric on $ M $ (not necessarily compatible with the covering map) for which all sectional curvatures of $ M $ are constant, equal to 1?

REPLY [7 votes]: Updated: If $n = 3$, then the answer to your question is yes. This is essentially the Thurston Elliptization Conjecture which was settled by Perelman's proof of the Geometrization Conjecture.
In my initial answer, I had hoped to use the obstruction $\alpha$ to construct a manifold which was finitely covered by $S^n$ which did not even admit positive scalar curvature metrics. As is discussed in the comments, this construction cannot work for lens spaces. In fact, it cannot happen for $n \geq 5$ at all, as was shown in Fake spherical spaceforms of constant positive scalar curvature by Kwasik & Schultz. 
In Paul Seigel's answer, he mentions that the answer to your question is no. Concretely, in Free Metacyclic Group Actions on Homotopy Spheres, Petrie shows that there is a smooth free action of $\mathbb{Z}_p\rtimes\mathbb{Z}_q$  on $S^{2q-1}$ where $p$ is odd and $q$ an odd prime. Such a quotient cannot admit a constant positive sectional curvature metric. However, by the result of Kwasik & Schultz, it does admit a constant positive scalar curvature metric.

Allow me to expand a little bit on the first part on Paul Siegel's answer before heading in a slightly different direction in the search for examples.
There is a homomorphism 
$$\alpha : \Omega_n^{\text{spin}} \to KO(S^n) = \begin{cases}\mathbb{Z} & n \equiv 0\bmod 4\\
\mathbb{Z}_2 & n \equiv 1, 2 \bmod 8\\
0 & \text{otherwise}\end{cases}$$
defined by Milnor. It was shown by Milnor and Adams that when $n = 1, 2 \bmod 8$ and $n \neq 1, 2$, there is an exotic $n$-sphere $\Sigma$ such that $\alpha(\Sigma) \neq 0$. In all other dimensions, exotic spheres have value zero.
It was later shown by Hitchin that if a closed spin manifold $X$ admits a metric of positive scalar curvature, then $\alpha(X) = 0$.
So, if $X$ is an $n$-dimensional closed spin manifold which admits a metric of positive scalar curvature and $n \equiv 1, 2 \bmod 8$, $n \neq 1, 2$, then $X\#\Sigma$ does not admit a metric of positive scalar curvature (connected sum is the addition operation in $\Omega^{\text{spin}}_n$ and $\alpha$ is a homomorphism).
If $S^n \to X$ is the $k$-fold universal cover (here $S^n$ denotes the standard smooth sphere), then the universal cover of $X\#\Sigma$ is $S^n\# k\Sigma$. The group of exotic spheres is finite (except possibly in dimension four), so let $d$ be the order of $\Sigma$. If $d \mid k$, then $S^n\# k\Sigma = S^n$ and you'd have an example of the phenomena you're looking for.
Note, this doesn't help when $n \equiv 2 \bmod 8$ as $S^n$ only covers $\mathbb{RP}^n$ which is not orientable, and hence not spin. When $n \equiv 1 \bmod 8$, then $S^n$ is the universal cover of infinitely many manifolds, e.g. lens spaces. In particular, for any $k$, there is an closed orientable manifold $X$ which has $S^n$ as its $k$-sheeted cover, all that's left to know is when such an $X$ can be chosen to be spin. I don't know how to do this off the top of my head, but the answers to this question seem promising. 
I believe that examples can be constructed this way, but I have never taken the time to construct one, sorry.