What exactly is a bound state and why does it have negative energy?
If you have a copy of Griffiths, he has a nice discussion of this in the delta function potential section. In summary, if the energy is less than the potential at $-\infty$ and $+\infty$, then it is a bound state, and the spectrum will be discrete: $$ \Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right). $$ Otherwise (if the energy is greater than the potential at $-\infty$ or $+\infty$), it is a scattering state, and the spectrum will be continuous: $$ \Psi\left(x,t\right) = \int dk \ c\left(k\right) \Psi_k\left(x,t\right). $$ For a potential like the infinite square well or harmonic oscillator, the potential goes to $+\infty$ at $\pm \infty$, so there are only bound states.
For a free particle ($V=0$), the energy can never be less than the potential anywhere***, so there are only scattering states.
For the hydrogen atom, $V\left(r\right) = - a / r$ with $a > 0$, so there are bound states for $E < 0$ and scattering states for $E>0$.
Update
*** @Alex asked a couple questions in the comments about why $E>0$ for a free particle, so I thought I'd expand on this point.
If you rearrange the time independent Schrödinger equation as $$ \psi''= \frac{2m}{\hbar^2} \left(V-E\right) \psi $$ you see that $\psi''$ and $\psi$ would have the same sign for all $x$ if $E < V_{min}$, and $\psi$ would not be normalizable (can't go to $0$ at $\pm\infty$).
But why do we discount the $E<V_{min}=0$ solutions for this reason, yet keep the $E>0$ solutions, $\psi = e^{ikx}$, when they too aren't normalizable?
The answer is to consider the normalization of the total wave function at $t=0$, using the fact that if a wave function is normalized at $t=0$, it will stay normalized for all time (see argument starting at equation 147 here):
$$ \left<\Psi | \Psi\right> = \int dx \ \Psi^*\left(x,0\right) \Psi\left(x,0\right) = \int dk' \int dk \ c^*\left(k'\right) c\left(k\right) \left[\int dx \ \psi^*_{k'}\left(x\right) \psi_k\left(x\right)\right] $$
For $E>0$, $\psi_k\left(x\right) = e^{ikx}$ where $k^2 = 2 m E / \hbar^2$, and the $x$ integral in square brackets is $2\pi\delta\left(k-k'\right)$, so
$$ \left<\Psi | \Psi\right> = 2\pi \int dk \ \left|c\left(k\right)\right|^2 $$ which can equal $1$ for a suitable choice of $c\left(k\right)$.
For $E<0$, $\psi_k\left(x\right) = e^{kx}$ where $k^2 = - 2 m E / \hbar^2$, and the $x$ integral in square brackets diverges, so $\left<\Psi | \Psi\right>$ cannot equal $1$.
It means the same thing it means in classical mechanics: if it is energetically forbidden to separate to arbitrarily large distance they are "bound".
The Earth is gravitationally bound to the Sun and the Moon to the Earth. Electrons in a neutral atom are electomagnetically bound to the nucleus. A pea rolling around in the bottom of a bowl is bound.
By contrast the Voyager probes are (barely) unbound and will fly (slowly) off into the galaxy.
Mathematically, bound states are states that decay sufficiently fast at infinity, so that the probability of finding the particle they describe in far away regions of space is negligible.
It has long been conjectured, based on physical intuition, it is the case for the meaningful quantum mechanical states, such as the eigenfunctions of the Hamiltonian (it is not expected that an atomic electron has a sensible probability of being at infinite distance from its nucleus).
This has been proved mathematically in the eighties, mainly by S.Agmon. Roughly speaking, the result is the following: eigenfunctions of the Schrödinger operator (i.e. corresponding to the discrete spectrum) are exponentially decaying in space. So if $\psi_n(x)$ are such eigenfunctions, $\lvert \psi_n(x)\rvert\leq A e^{-B\lvert x\rvert}$, for some positive constants $A,B$.