Let's take a simple original picture to look at - just two nearby dots on a white background. If you have bad vision, the dots look blurred.
The way good vision works is to ensure that all the light hitting any particular small area of your retina comes from the same direction in front of you. Conversely, all the light coming from one direction hits one specific spot on your retina.
When you have bad vision, the light from a locus of nearby directions all hits on the same part of your retina, and the light from a particular direction is smeared out over an area on your retina. Hence, blurred vision is an averaging effect. When you look at the dots, you'll see them smear out into each other.
You might try to compensate for this by making a "counter-blurred" image where the source dots are smaller, but if the original dots are close enough that light from the center of one dot is spilling over to overlap light from the center of the second dot, making the dots smaller won't fix that problem. Hence, the dots will always appear blurred. You can't create the impression that the original has for someone with good vision.
A photograph is really just a bunch of nearby dots, and so the same problem applies.
I don't know about the 3D monitor, though. I suppose if it can control the direction of light coming off it, it could be modified to focus the light some and create a sharp image for someone with blurred vision.
Dear Rootosaurus, when you're looking at an image of a chair behind you in a flat mirror, then you're observing the so-called virtual image of the chair. If the mirror's surface is located in the $x=0$ plane and the coordinate of the real chair is at $(x,y,z)$, then the virtual image of the chair is at $(-x,y,z)$.
However, the light rays coming from the real chair that are reflected by the mirror and that ultimately end up in your eyes have exactly the same directions as the light rays of a hypothetical chair that would be actually located at the point $(-x,y,z)$ behind the mirror. So geometrically, you can't really distinguish a reflection of a chair that seems to be located at $(-x,y,z)$ from a real chair that is located at $(-x,y,z)$ and that you're observing through a window without a mirror. The geometry of the light rays is identical. That's why the concept of images is so useful.
In particular, myopia means that one has some trouble to observe distant objects. Distant objects - imagine a distant point - have the property that they emit light rays that are nearly parallel. The further an object is, the more parallel its light rays look when they arrive to your eye. However, the lenses in your eyes have to convert these parallel mirrors into converging mirrors - so that all the light rays coming from the distant star end up at one point of the retina.
Myopic eyes are good in converting "steeply divergent" light rays from nearby objects to converging ones, but they're doing too much of a good thing. When you get too parallel light rays, myopic eyes make them converge too much, too early - the intersection will be inside the liquid in your eye. Hyperopia is the opposite disorder in which eyes make the light converge less than is needed. But what's relevant for your question is that the virtual image of the chair at $(-x,y,z)$ "emits" the same excessively parallel rays as a real chair at the same point, so a myopic eye will have the same trouble seeing it. After all, it shouldn't be paradoxical: the total distance that the light rays have traveled includes the distance of you from the mirror as well as the distance of the mirror from the real chair - because the colorful photons ultimately came from the chair and they were just reflected, not created, at the plane of the mirror.
Best Answer
Fun question!
As you pointed out,
$$\theta \approx 1.22\frac{\lambda}{D}$$
For a human-like eye, which has a maximum pupil diameter of about $9\ \mathrm{mm}$ and choosing the shortest wavelength in the visible spectrum of about $390\ \mathrm{nm}$, the angular resolution works out to about $5.3\times10^{-5}$ (radians, of course). At a distance of $24\ \mathrm{km}$, this corresponds to a linear resolution ($\theta d$, where $d$ is the distance) of about $1.2\ \mathrm m$. So counting mounted riders seems plausible since they are probably separated by one to a few times this resolution. Comparing their heights which are on the order of the resolution would be more difficult, but might still be possible with dithering. Does Legolas perhaps wiggle his head around a lot while he's counting? Dithering only helps when the image sampling (in this case, by elven photoreceptors) is worse than the resolution of the optics. Human eyes apparently have an equivalent pixel spacing of something like a few tenths of an arcminute, while the diffraction-limited resolution is about a tenth of an arcminute, so dithering or some other technique would be necessary to take full advantage of the optics.
An interferometer has an angular resolution equal to a telescope with a diameter equal to the separation between the two most widely separated detectors. Legolas has two detectors (eyeballs) separated by about 10 times the diameter of his pupils, $75\ \mathrm{mm}$ or so at most. This would give him a linear resolution of about $15\ \mathrm{cm}$ at a distance of $24\ \mathrm{km}$, probably sufficient to compare the heights of mounted riders.
However, interferometry is a bit more complicated than that. With only two detectors and a single fixed separation, only features with angular separations equal to the resolution are resolved, and direction is important as well. If Legolas' eyes are oriented horizontally, he won't be able to resolve structure in the vertical direction using interferometric techniques. So he'd at the very least need to tilt his head sideways, and probably also jiggle it around a lot (including some rotation) again to get decent sampling of different baseline orientations. Still, it seems like with a sufficiently sophisticated processor (elf brain?) he could achieve the reported observation.
Luboš Motl points out some other possible difficulties with interferometry in his answer, primarily that the combination of a polychromatic source and a detector spacing many times larger than the observed wavelength lead to no correlation in the phase of the light entering the two detectors. While true, Legolas may be able to get around this if his eyes (specifically the photoreceptors) are sufficiently sophisticated so as to act as a simultaneous high-resolution imaging spectrometer or integral field spectrograph and interferometer. This way he could pick out signals of a given wavelength and use them in his interferometric processing.
A couple of the other answers and comments mention the potential difficulty drawing a sight line to a point $24\rm km$ away due to the curvature of the Earth. As has been pointed out, Legolas just needs to have an advantage in elevation of about $90\ \mathrm m$ (the radial distance from a circle $6400\ \mathrm{km}$ in radius to a tangent $24\ \mathrm{km}$ along the circumference; Middle-Earth is apparently about Earth-sized, or may be Earth in the past, though I can't really nail this down with a canonical source after a quick search). He doesn't need to be on a mountaintop or anything, so it seems reasonable to just assume that the geography allows a line of sight.
Finally a bit about "clean air". In astronomy (if you haven't guessed my field yet, now you know.) we refer to distortions caused by the atmosphere as "seeing". Seeing is often measured in arcseconds ($3600'' = 60' = 1^\circ$), referring to the limit imposed on angular resolution by atmospheric distortions. The best seeing, achieved from mountaintops in perfect conditions, is about $1''$, or in radians $4.8\times10^{-6}$. This is about the same angular resolution as Legolas' amazing interferometric eyes. I'm not sure what seeing would be like horizontally across a distance of $24\ \mathrm{km}$. On the one hand there is a lot more air than looking up vertically; the atmosphere is thicker than $24\ \mathrm{km}$ but its density drops rapidly with altitude. On the other hand the relatively uniform density and temperature at fixed altitude would cause less variation in refractive index than in the vertical direction, which might improve seeing. If I had to guess, I'd say that for very still air at uniform temperature he might get seeing as good as $1\rm arcsec$, but with more realistic conditions with the Sun shining, mirage-like effects probably take over limiting the resolution that Legolas can achieve.