Pressure of air in a cup If I have a 1.0m long cylindrical cup, and fill it up to $h$ = 0.5m with water, then the 0.5m of air above it will be at atmospheric pressure. But if I now place a perfect seal on the cylinder, and then invert the cylinder, what is the pressure of the air itself in the cylinder? There is now no air molecules above the 0.5m of air in the cylinder, and thus no net force downwards due to all the air molecules from the atmosphere. Thus there should no longer be this large net pressure on the air (i.e. the air in the cylinder is no longer at atmospheric pressure), and I would assume the only pressure within the cup at the seal is given by $p = p_{water} + p_{air} = \rho_{water}gh + nRT/V$ where $n$ is the density of air in the cylinder, $T$ is the temperature, and $V$ is $(1.0 - h)A_{cylinder}$ (the latter is the area of the cylinder). 
This is based on a similar post. 
 A: 
then the 0.5m of air above it will be at atmospheric pressure.

Correct.  You don't give the width of the cylinder, so the exact height isn't that relevant.  But we can say there is a volume of air $V$ remaining in the cylinder.

But if I now place a perfect seal on the cylinder, and then invert the cylinder, what is the pressure of the air itself in the cylinder?

As the volume has not changed (and we presume the temperature is also constant), then $P = \frac{nRT}{V}$, so the pressure is the same as it was before.

There is now no air molecules above the 0.5m of air in the cylinder, and thus no net force downwards due to all the air molecules from the atmosphere.

That doesn't matter.  The vessel is rigid.  The air molecules inside bump against the walls.  The walls respond with a normal force.  That force over their area is a pressure.  The pressure is the same as the atmosphere because it started at that pressure.  Assuming it was strong enough, you could take it into outer space and the same pressure would remain on the inside.
The water is just a complication.  It has nearly constant volume, so the air is constrained to a nearly constant volume in the vessel.  The same would happen if you sealed it without water inside.

why can cardboard (or any sufficiently low-mass sealant) remain stuck to the cup?

Because the cardboard doesn't have to drop very far to equalize the pressure.  This is normally done with a cup.  Let's assume a cup 8cm tall, with 4cm of water in it.
When we turn it upside down, what pressure differential will hold the water in place?  That would be equivalent to the static pressure at the bottom of the water.
$$P = \sigma g h$$
$$P = (\text{g}/\text{cm}^3)(9.81\text{m}/\text{s}^2)(4\text{cm}) = 392\text{Pa}^2$$
So what volume change lowers the atmospheric pressure that much?
$$PV = P'V'$$
$$V' = \frac{P}{P'}V$$
$$V' = \frac{101325\text{Pa}}{100933\text{Pa}}V$$
$$V' = 1.0039 V$$
Since the sides are rigid, the entire volume change is accommodated by a change in height.  
$$h' = 1.0039h = 4.016\text{cm} = h + 0.16\text{mm}$$
So the pressure change is met by lowering a bit more than a tenth of a millimeter.  As we can see water remain in a straw that is several millimeters wide, that's no problem for the water to remain in place.
Really, the limiting case will be where one side of the card remains tight against the cup, while the other side drops to take the entire change.  On average this will double the height that might appear, so assume you really cause it to drop by a third of a millimeter.  Still not a problem.  

Your calculation neglected the mass of the sealant (e.g. if it's 10kg).

Yes, I'm assuming a light card that can be neglected.  The calculation just needs to add in the cross-sectional density of that material if it's significant.  You can certainly pick materials that are light enough to ignore.

But the cup's cross-section does matter since it is the pressure difference acting across this cross-section on the sealant that support it from falling against gravity.

Nope.  The wider the cup, the heavier the water, but also the greater the force from the atmosphere.  These two effects both scale with the cross section so there's no change with area.  That's why I didn't have to specify a particular area for the cup, just know the volumetric density of water and the atmospheric pressure.  
