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.
Here is a heuristic. The actual details will depend on the details of what type of rock it is, and materials science and chemistry beyond my pay grade, but this gives what I think should be the general idea. 

All rocks get wet when you put water on them, the surface gets slick, and the like. When this happens, what you get is the water adhering to the surface of the rock. Let's say that if the surface area of the rock is $A$, then the mass of water that adheres to the rock is $\sigma A$, where $\sigma$ is some number that depends on the material that the rock is made out of, how porous the surface is, etc, etc.

Now, we have a rock that has water mass $\rho\_{w}\sigma A$ and rock mass $\rho\_{r}V$. Thus, our wet rock has a water content fraction of $\frac{\rho\_{w}\sigma 4\pi r^{2}}{\rho\_{r}\frac{4}{3}\pi r^{3}} = 3\sigma\frac{\rho\_{w}}{r\rho\_{r}}$. So, for large rocks, the rock is mostly made out of rock, but for small rocks, the rock is mostly made out of water. Thus, as the particle size decreases, we expect the rock to behave in more water-like ways, because the rock is more and more made out of water. 

In reality, this will be further magnified because the water will extend out from the rock some, and the particle-particle interactions will depend on things like surface tension and the like (not to mention that there will be some cutoff to all of this, because you don't expect the ratio to ever diverge). But the per-particle water/rock ratio alone shows you that you expect the thing to behave in a more water-like way as the particle size decreases.