Question: I think of sand as a lot of very small rocks.

Suppose I have a pile of rocks, each about 1cm in size, and the pile is a meter tall. If I pour a bucket of water on the rocks, most of the water will fall through the rocks and form a puddle on the ground.

On the other hand, if I have a pile of sand one meter tall and I pour a bucket of water on it, the water will stick to the sand and I'll have a bunch of wet sand, and not much water on the ground.

Further, the wet sand acts distinctly differently from dry sand. For example, the angle of repose of wet sand is different from dry sand. An hourglass with wet sand it in might not work, or would run at a different speed from a dry hourglass. But I don't think the angle of repose of a pile of wet 1cm rocks is significantly different from that of dry rocks.

So, how small should rocks be before they show wetting properties the way sand does?

Answer: 
For a packing of grains to stay wet up to a height $h$, the gravitational pressure $\rho g h$ needs to be balanced by the capillary pressure $\sigma cos(\theta)/r$. Here, $r$ represents the effective pore radius of the packing, $\theta$ the wetting angle (angle at which the air-water interface meets the sand grains), $\rho$ the water density, $g$ the gravitational acceleration, and $\sigma$ the surface tension associated with the air-water interface.

It follows that the hydrostatic head $h$ is given by:
$$h=\frac{\sigma cos(\theta)}{\rho g r}$$
For an order-of-magnitude estimate you can equate the capillary radius $r$ to the grain radius $R\_{grain}$, and $cos(\theta)$ to unity (fully water-wet grains). Together with $\sigma \approx 0.1 N/m$, $\rho \approx 10^3 kg/m^3$, and $g \approx 10 m/s^2$, it follows that
$$h \approx \frac{A\_{cap}}{R\_{grain}}$$
with $A\_{cap} \approx 10^{-5}m^2$.

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As an aside, petroleum engineers use an approach like the above (replacing the air by oil) to determine the transition zone thickness in oil reservoirs. This is the zone over which the oil saturation 'builds up' above the water-bearing zone. As reservoir rock is usually fine-grained, and $\rho$ is replaced by the oil-water density difference, this transition zone can be tens of meters thick.
Wetting here is most likely a capillary effect: Your question is about the size of the air gaps between grains (or rocks) of sand, not the size of the grains or rocks. In practice, except perhaps for very peculiarly shaped objects, these will be of similar magnitude to the smallest gap-filling grains.

What happens is that the energy required to create the air-liquid surface tension is not needed where the water does not have an air interface but contacts a material. With glass (or sand), there is a strong attraction between the polar water molecules and the polar silicon oxide material. Hence the water gets (slightly) sucked into especially small gaps. For this [capillary action](http://en.wikipedia.org/wiki/Capillary_action) to cause pure water at room temperature to rise one meter in a glass capillary, the diameter would have to be at most $15\,\mathrm{\mu m}$. That is a (very rough) measure for the gap size you can have without water starting to partially drain. Note that this draining would be gradual: Only the (few?) gaps that are too large and too high would loose their water. Hence the observation that the sand is somewhat wet may still occur even if only some paths of the smallest gaps suffice for capillary wetting.

There is a good chance that your grains of sand (or rocks) may be porous and hence wet at least a bit better than you would expect from similarly sized pieces of unfractured glass. This will not matter much for the pile of rocks (so each rock gets slightly wetter/soaked), but might make all the difference for a pile of sand where internal capillaries might allow sand to soak up water even if all other paths of actual gaps were too large to allow it. Taken to the extreme, for example in silica gel, it allows sand-like grains to strongly adsorb water even just as humidity.