Must a manifold covered by $ S^n $ admit a metric of constant positive sectional curvature? 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?
 A: 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.
A: Hitchin showed (here, I think) that there is an exotic sphere which admits no metric of positive scalar curvature.  This manifold certainly has the sphere as its topological universal cover, but its sectional curvatures can't all be positive let alone 1.

The OP is also interested in the smooth case.  I think there still are counterexamples, sometimes called "fake lens spaces" - this is a useful term to google, though many of the hits focus on the topological category.
The first step is to classify spherical space forms (i.e. manifolds with constant sectional curvature 1).  By various classical results in Riemannian geometry this amounts to classifying all finite groups which act freely on the sphere by isometries.  This has been done, and so your question becomes "are there any other finite groups which act freely and smoothly on the sphere?"  The answer is "no" in even dimensions because the only nontrivial group which can act on an even dimensional sphere is $\mathbb{Z}/2\mathbb{Z}$ (the Euler characteristic is multiplicative for coverings). The answer is also "no" in dimension 3 by the geometrization theorem.
But more interesting things can happen in higher dimensions.  The results inevitably use a lot of heavy duty surgery theory, so I'll point you to a survey paper by Hambleton which goes into more detail.  He credits Lee and Petrie for producing the first smooth counterexamples.
