Can vacuum breakdown occur from a positively charged surface? Against a sufficiently large voltage, resistance is futile.
Although the vacuum is a very good insulator, electrical breakdown can occur even in a perfect vacuum in the presence of a very strong electric field.  Typically, this is explained by electrons escaping from a negatively charged electrode, through Fowler-Nordheim tunneling, and flowing to a positively charged electrode.  (This topic has been discussed several times on this site, e.g. How can a vacuum have a breakdown voltage? and Is lightning possible/visible in vacuum or not?).
Is it possible for vacuum breakdown to occur starting from a positively charged electrode?  For example, imagine a small spherical object with an extremely high positive charge.  Would positive ions start to escape from the surface?  If so, what surface field strength would be required to trigger an electric discharge?
 A: Yes, sufficiently high voltage will cause ions to leave a positive electrode.
You can reach that conclusion with just a thought experiment: If you removed all the electrons from the electrode (i.e. the maximum theoretical voltage), it would immediately vaporize, since the electric fields pushing its atoms apart would greatly overcome the gravitational fields holding them together.  So, there must be a balance point where enough electrons have been removed to start this process, and begin emitting ions from the surface.
The deeper question, though, is "Does this ion emission lead to a breakdown?" and that is a different matter.  To cause a breakdown, the emitted ion would need to hit something, and liberate electrons, which fly back to the electrode surface, spall off more ions, those fly out and hit something else, and then the process repeats in a chain reaction until the electrode is discharged.  For this chain reaction to happen, though, you must have something for the ion to hit.  If the ion just flies off to infinity, you don't have breakdown, you have the electrode slowly boiling itself away into space.
This is a key limit of vacuum chamber research: Every vacuum chamber has walls.  So, eventually, every ion that boils off hits those walls, and bam, there is a breakdown.  In practice, this happens well before you get to ion boil-off voltages, because cosmic rays can penetrate the chamber walls, excite an electron from anything they hit, and in a perfect vacuum the electron has a mean free path so long that it hits the electrode at a very significant fraction of light speed, and can cause ions to spall off and start the breakdown early. Even a perfect vacuum chamber still suffers this vulnerability.
The result is that very high field electrostatic generators (like pelletrons) enclose the electrodes in pressurized gas (usually sulfur hexafluoride) instead of vacuum. The gas's dielectric strength is technically lower, but the mean free path of emitted particles is so short that they can't get up enough speed to ionize the pressurized gas, nor to spall material from the walls or electrode.  They basically bump into too many gas molecules to ever get going very fast.  Even under these conditions, electrode charges still max out around 25-30 megavolts at practical sizes, beyond which corona losses exceed the ability of the charging systems to compensate for them, so charge bleeds away faster than it can be added.
What if you tried it in orbit, though?  No walls, and the mean free path of a particle can exceed 50 km.  That means by the time your ion hits even one other atom, it will be 50 km away from your electrode; any electrons it frees in the collision will be more likely absorbed by other local ions than attracted all the way back to your electrode. So, in theory, there is no reason you couldn't achieve much higher voltages than anything we have ever seen in a lab.  Under such conditions, you might be able to trigger ion boil off without ever having a distinct breakdown point.
