Why H3O+ exists and H3O- doesn't?
I'm not a big fan of MOs when this can be explained using just AOs. For sure, MO theory gives a much better and profound explanation, and I'm surely waiting for someone to do that, but that's what I would answer if the kids asked me this.

Oxygen has 6 electrons on its outer shell, which has $\mathrm{2s}$ and $\mathrm{2p}$ orbitals. The arrangement of these electrons is such that $\mathrm{2s}$ sublevel is completed, and $\mathrm{2p}$ sublevel contains 4 electrons. Following Hund's rule,  two electrons occupy one of three $\mathrm{2p}$ orbitals and other two come by one to each of $\mathrm{2p}$ orbitals left.



When water forms, oxygen shares two non-bonded electrons with hydrogen atoms, forming a covalent bond with them. 

$\ce{H3O+}$ forms via donor-acceptor bonding, when oxygen supplies its lone pair, which it has two, to form a bond with $\ce{H+}$ ion. 

To form $\ce{H3O-}$, you'll need an interaction between water molecule and $\ce{H-}$ ion. Even if we ignore the fact that hydride is a powerful base ($\mathrm pK_\mathrm a$ is around $42$) and will react violently with water, it would be impossible. In this theoretical case neither of the species has an orbital suitable to accept a lone pair from another species. Each has a complete outer shell.

Edit: it really exists though. Wow.

Sources: 


Probing the transition state via photoelectron and photodetachment spectroscopy of H3O−
Calculation of the photodetachment spectrum for H3O−