The hypothesis $K\leq 0$ in the proof of Hadamard's Theorem In chapter 7 from do Carmo's Riemannian Geometry, right after proving Hadamard's theorem, there is the following remark:

When he says "poles can exist in non-compact manifolds which have positive sectional curvature", I believe he is trying to justify the hypothesis "$K\leq 0$".
What I don't understand is: why did he have to talk about "non-compact manifolds"?
I thought he would say "poles can exist in complete manifolds with positive curvature", which makes perfect sense and justifies the hypothesis "$K\leq 0$". 
I know I'm not crazy because exercise 13 (where $M$ is the parabolloid $z=x^2+y^2$) is precisely an example of a complete manifold with positive curvature having a pole (namely, the origin).
So what is this "non-compact" business?
 A: The reason Do Carmo restricts that comment to noncompact manifolds is because of some results that he will prove later, which show among other things that a manifold with sectional curvatures bounded below by a positive constant must have a conjugate point along every geodesic. In particular, as @MoisheCohen commented, if $M$ is compact and has strictly positive sectional curvature, then the curvature has a positive lower bound and thus $M$ has no poles.
But I don't understand your remark that the existence of poles in the positive curvature case would "justify the hypothesis $K\le 0$." My interpretation of that comment is sort of the opposite -- Hadamard's theorem shows that for complete manifolds, $K\le 0 $ is sufficient to guarantee the existence of poles, so one might wonder whether it's also necessary. Do Carmo's comment and the paraboloid example are meant to demonstrate that $K\le 0$ is far from necessary, in the sense that it's even possible to find a complete manifold with $K>0$ everywhere and that nonetheless has a pole.  
It's worth noting, by the way, that Do Carmo's definition of a pole is nonstandard. Most Riemannian geometers define a pole to be a point $p\in M$ such that the restricted exponential map $\exp_p\colon T_pM \to M$ is a global diffeomorphism. That implies that $p$ has no conjugate points, but it is much stronger. Of course, using the more usual definition of a pole, it's immediate that a manifold with a pole has to be diffeomorphic to $\mathbb R^n$, so it can never be compact.
