Can't a hydrogen electron have net positive energy? This page
http://www.physicspages.com/2011/06/07/hydrogen-atom-series-solution/
is the 2nd half of a solution to the hydrogen atom Schrödinger equation.
They derive that $E = -1/n^2 *$ (bunch of positive constants)  (equation 29).
So this seems to be saying that the energy of the electron,
that is the net of potential energy plus kinetic energy, must
be negative.
But couldn't a hydrogen electron be excited somehow so that its
kinetic energy outweighs its potential energy  (its potential
energy is negative since infinite separation of electron from
nucleus is defined to be 0 potential energy)  and thus have a
net positive energy?
 A: The energy is defined as 
$$ E = \frac{p^2}{2m} + V(\vec r) $$
where the first term is the kinetic energy and the second term is the potential energy calibrated so that $V(\vec r)=0$ for $|\vec r|\to\infty$. 
Consequently, you may say that the energy in a given state (an analogy of an orbit in classical physics) is equal to the kinetic energy $T_\infty$ that the electron has when it escapes to infinity, $|\vec r|\to\infty$.
When this $T_\infty$ is positive, it means that the electron may – and almost certainly will – escape to infinity and never return back, approaching some constant nonzero velocity $\vec v$ as it flies away from the nucleus. If the electron escapes in this way, we say that the atom has been ionized. It is no longer a bound state. It doesn't make sense to call it a hydrogen atom. The corresponding classical trajectories would be hyperbolae, not ellipses as they are for the bound states.
In other words, for $E=T_\infty \gt 0$, we have the system of the electron and the nucleus, but they are not bound and shouldn't be described as "one object". The spectrum of this system – a part of the spectrum that one gets from solving the equations for the "hydrogen atom" – is continuous. This is characteristic for a composite particle that has been separated to pieces.
We need $E=T_\infty \lt 0$ which guarantees that the particle doesn't have enough energy to escape to $|\vec r|\to \infty$ at all. Consequently, the probability is basically 100 percent that the electron (in this case) is closer than a millimeter from the nucleus. They're bound. They're a bound state. One may effectively say that the electron is confined (by the potential energy profile and by its limited extra kinetic energy) to a finite region around the hydrogen atom. 
If that's so, if we have a bound state, the energy is negative but the spectrum is discrete – much like it is discrete for a particle in a potential well or any other "compact" space.
So the spectrum of the Hamiltonian for the hydrogen atom is "mixed". It has the discrete states with $E\lt 0$ that you mentioned and that described an actual atom, a bound state of the nucleus and the electron. And then the spectrum has a continuous part for $E\gt 0$. For all values $E\gt 0$, there exists an eigenstate and it's some deformed plane wave for the electron describing the electron moving from space mostly away from the nucleus.
There are also eigenstates for $E=0$ (the marginal case corresponding to the parabolic trajectories in classical physics) but they're a measure-zero "edge" of the continuous spectrum – or, equivalently, an $n\to \infty$ limit of the discrete bound states – and the subtleties with the exact value $E=0$ may be pretty much neglected.
