Suppose that the smooth manifold $ M $ has the nsphere 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?
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.

6$\begingroup$ It is not polite to change the question after a good answer. I suggest that you revert to the original, accept a good answer to your question, and ask a new question. $\endgroup$ – Lee Mosher Jul 1 '18 at 20:56

2$\begingroup$ @LeeMosher: You are right. I thought about this but decided to edit the question all the same because Paul had replied almost immediately after the question had been posted, and because the edit consisted of 1 word only. I have followed your suggestion to revert to the original question and accept his answer which is perfectly adequate for the original question. I will however not post a new question as it is too similar to this one. I will leave the question that I had intended as a comment (see below). I apologize to everyone for my mistake in the original statement. $\endgroup$ – rgnrmllbrg Jul 1 '18 at 22:16

$\begingroup$ @PaulSiegel: Thanks for your reply. Do you know a counterexample when the covering map is required to be smooth (not just continuous)? This is what I had intended to ask originally, as I didn't expect any subtlety involving exotic spheres. $\endgroup$ – rgnrmllbrg Jul 1 '18 at 22:19

2$\begingroup$ @rgnrmllbrg I added some remarks on the smooth case to my answer  I believe the conclusion is that there are still counterexamples, but they use a lot of surgery theory and so I can't explain them with much certainty. $\endgroup$ – Paul Siegel Jul 1 '18 at 22:56

$\begingroup$ @PaulSiegel: Thank you very much for your thorough reply, I really appreciate it! I will look at the paper that you mentioned, but I already got what I need from what you wrote. $\endgroup$ – rgnrmllbrg Jul 1 '18 at 23:02
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.

$\begingroup$ I haven't checked all the details, but I'm not sure this approach works. First, from Browder's result on the Kervaire invariant, $bP_{n+1}$ has order $2$ whenever $n$ is $1$ mod $8$. This implies $\theta_n = d$ is even in these dimensions (other than $n=1$), so $k$ must be even. Now, for any of the usual lens spaces, this means the covering $S^n\rightarrow L$ factors through $S^n\rightarrow \mathbb{R}P^n$. But in dimension $1$ mod $8$, $\mathbb{R}P^n$ is not spin. This implies none of these lens spaces are spin. (Of course, $S^n$ could cover other things....) $\endgroup$ – Jason DeVito Jul 17 '18 at 18:03