How can the ratio of a black hole's (or any object's) angular momentum to its mass be less than one? In the latest Quanta Magazine article, about Kerr black holes, the researchers say their new solution for these objects only works, so far, for Kerr black holes with a ratio of angular-momentum-to-mass of less than one...
How is this possible?  If angular momentum is velocity times mass, how can that number be less than just the mass?
 A: For one thing, angular momentum has units of $\mathrm {length*mass*velocity}$. But they are likely referring to the Kerr Metric for a rotating body. There is a length parameter $a=J/Mc$ ( J angular momentum, M mass, c speed of light) that is a parameter used to describe Kerr black holes. A unit system where $c=1$ is often used, which would leave $a=J/M$.  Saying J is less than M implies $a<1$.
A: To answer this we need to go over a few preliminaries. A rotating uncharged black hole has two event horizons with radii (using SI units):
$$ r = \frac{G}{c^2} \left(M \pm \sqrt{M^2 - \left(\frac{c^2}{G}a\right)^2}\right) $$
where $a$ is known as the Kerr parameter and is given by:
$$ a = \frac{J}{Mc} $$
Here $J$ and $M$ are the angular momentum and mass as measured by an observer at infinity. The physical meaning of the angular momentum and mass are a little subtle given that the Kerr black hole is a vacuum solution and contains no mass, but we don't need to worry about this for the purposes of this discussion.
To avoid multiple factors of $G$ and $c$ cluttering up our equations we usually use geometric units that have $G = c = 1$, and this tidies up the equations considerably to:
$$ r = M \pm \sqrt{M^2 - a^2} $$
$$ a = \frac{J}{M} $$
If the black hole is not spinning at all the angular momentum is $J = 0$ and therefore the Kerr parameter is $a=0$ so we find the two horizons are at $r = 2M$ and $r = 0$, and $r=2M$ is the usual event horizon radius for a static black hole. So far so good.
As we increase the angular momentum the inner horizon grows outwards from $r=0$ and the outer horizon shrinks inwards from $r=2M$, and when $a$ reaches $a=1$ the two horizons meet and merge to form an extremal black hole. This is a very strange object and thought to be physically unreasonable. It has a naked singularity that causes various problems including a breakdown of causality, and this geometry isn't well understood.
The statement in the Quanta article:

So far, stability has only been proved for slowly rotating black holes — where the ratio of the black hole’s angular momentum to its mass is much less than 1.

is referring to the value of the Kerr parameter, $a$, and it means the proof discussed in the article is valid only for rotating black holes that are far from extremal i.e. have $a \ll 1$ so they are rotating much slower than an extremal black hole of the same mass. This is not a surprising limitation given how weird extremal black holes are and how poorly they are understood.
So to (finally) get to your specific question:

How is this possible? If angular momentum is velocity times mass, how can that number be less than just the mass?

The angular velocity can be zero simply by the black hole not rotating, and in that case the angular momentum is $J=0$ so the Kerr parameter $a=0$.
