How do holes move across a PN junction? When learning about the formation of the depletion region when n-type and p-type semiconductors are brought together, I note that many resources state that whilst electrons diffuse from the n-type material to the p-type material, holes also diffuse in the other direction.
The Feynman Lectures section 14-4 says this:

When the electrons in the n-type material arrive at the boundary they will not be reflected back as they would at a free surface, but are able to go right on into the p-type material. Some of the electrons of the n-type material will, therefore, tend to diffuse over into the p-type material where there are fewer electrons. This cannot go on forever because as we lose electrons from the n-side the net positive charge there increases until finally an electric voltage is built up which retards the diffusion of electrons into the p-side. In a similar way, the positive carriers of the p-type material can diffuse across the junction into the n-type material. When they do this they leave behind an excess of negative charge. 

, and Wikipedia says this:

Electrons and holes diffuse into regions with lower concentrations of them, much as ink diffuses into water until it is uniformly distributed. By definition, the N-type semiconductor has an excess of free electrons (in the conduction band) compared to the P-type semiconductor, and the P-type has an excess of holes (in the valence band) compared to the N-type. Therefore, when N-doped and P-doped semiconductors are placed together to form a junction, free electrons in the N-side conduction band migrate (diffuse) into the P-side conduction band, and holes in the P-side valence band migrate into the N-side valence band.

This idea of holes and electrons both diffusing across the junction is also shown in this video, and is shown in the following image:
                                                    

I am struggling to understand the idea of holes being able to move across the junction, however. My understanding is best shown by the HyperPhysics page on the PN junction, which says the following:

When a p-n junction is formed, some of the free electrons in the n-region diffuse across the junction and combine with holes to form negative ions. In so doing they leave behind positive ions at the donor impurity sites.

In my mind, the diffusion of electrons causes holes in the p-type region near the junction to be filled. Additionally, the departure of these electrons from the n-type region leaves a net positive charge behind - notably, however, despite this positive charge there should be no holes, as the donor atoms would retain a full outer shell. Thus, whilst this movement fills holes, it creates no holes in its place. This seems to put my interpretation (and seemingly HyperPhysics) at odds with the other mentioned resources. 
I would like to look at the following quotes from the Feynman Lectures:

This cannot go on forever because as we lose electrons from the n-side the net positive charge there increases until finally an electric voltage is built up which retards the diffusion of electrons into the p-side

, and:

In a similar way, the positive carriers of the p-type material can diffuse across the junction into the n-type material

The first quote seems to be referencing the behaviour described by HyperPhysics. The second quote, however, seems to be mentioning the diffusion of holes as independent to the diffusion of electrons.
My initial interpretation of these materials had me considering that the description of "holes" was being applied to the n-type side despite each atom having a full outer shell, simply as a tool for describing the behaviour. This, however, seems incompatible with the following Wikipedia quote:

Following transfer, the diffused electrons come into contact with holes and are eliminated by recombination in the P-side. Likewise, the diffused holes are recombined with free electrons so eliminated in the N-side.

So, what is it? Is the diffusion of electrons causing a net positive charge on the n-type side of the depletion region, or are holes diffusing independently across, or somehow both? If holes are indeed diffusing across, how do they manage to get past the diffusing electrons without being filled?
 A: 
I am struggling to understand the idea of holes being able to move across the junction, however.

Holes move in just the same way electrons do. 
One minor difference is that because the curvature of the E-k curve for the valence band is different from the curvature for the conduction band, holes and electrons have different effective masses. 
Other than that, they move the same way. If electrons can move across the junction, holes can do the same thing.
In comments you added,

As I understand it, the driving force behind electron diffusion is the repulsion of like charges. The holes are all in the valence band and so unlike electrons cannot move freely unaided, and thus I wouldn't expect them to naturally repel and move into the n-type region.

This is incorrect on a couple of points.
First, just like when uncharged molecules diffuse in a gas, the driving force behind diffusion is nothing more than that the particles are moving randomly, so if there are more of them at point A and fewer at a nearby point B, there will be a net flow of particles from A towards B. 
To the extent that the mutual repulsion of the electrons causes a current, we would call that part of the drift current, not part of the diffusion current. (Also remember that, for example, in the n-region there are vast numbers of electrons and they repel each other, there are also fixed positive charges at the ionized donor sites that attract the mobile electrons, and the associated fields can cancel each other out)
Second, electrons can't move freely in the valence band because nearly all electron states are full. But this doesn't prevent holes from moving in the valence band. Holes in the valence band move readily, because the valence band isn't jam-packed with holes, it's jam-packed with electrons. Holes in the valence band move nearly as readily as electrons in the conduction band (see the point about different effective masses above).
In your comment reply to boyfarrel, you said,

Would it be correct to say that the diffusion of the free electrons from the n-type region into the p-type region causes electrons from thermally generated pairs to do the same, thus causing holes to gradually progress further away from the p-type region (and so into the n-type)?

This is again not quite right. 
There are (relatively) many electrons in the n-region. These not only got excited from the donor sites into the conduction band, but thermodynamics says they'll take a certain distribution of energies above the conduction band edge. Because of this, a small fraction of them will have enough energy to overcome the potential barrier of the junction, should they randomly happen to move in that direction. 
Similarly, on the p-side there are relatively many holes, and a small fraction of these have sufficient energy to overcome the junction barrier should they randomly happen to move in that direction. 
The small fraction of the holes from the p-side that happen to cross the barrier over to the n-side form an excess population relative to the equilibrium hole population on the n-side. So, through random motion they have net motion away from the junction.
At the same time, as they randomly move around on the n-side, they have a probability to recombine with the vastly more numerous electrons present on the n-side. So we see (if the n-side is big enough) an exponential drop in this excess hole population as we move away from the junction. (the same way there's an excess electron population on the p-side, dropping exponentially as we move further into the p-side)
The p-side excess electrons aren't there because of anything the holes did, and the n-side excess holes aren't there because of anything the electrons did. They're there because there's a big honking crowd of electrons or holes on the other side of the junction, and a small fraction of those have managed to "jump the fence" and end up on the "wrong" side of the junction.
