From idle observation of a typical hourglass, the grains of sand on the top appear to slide towards the centre, and the position of that centre is itself ‘dropping’.
If you could position one red grain amongst all the other white grains (as a marker), where in the hourglass would be the best position to be the last grain to fall?
I realise there may be too many variables to prove a constant and specific answer.
I would think it is determined by the angle of repose.
If looking at a hour glass which is a cylinder for it’s body and then curves evenly to the vortex then a grain on the outer edge at the point the glass begins to curve would be at the the lowest angle from the vortex and greatest distance away along that angle .
Top surface, touching the glass, on the left.
Check this out: https://www.inf.utfsm.cl/~amoreira/orzelc.pdf
They have photos with colored grains so that you can see the process.
Spoiler: It depends on the shape of the hourglass. In hourglasses with shallow angles or curved shapes, a crater forms and expands outwards, so the grains touching the glass near the aperture are the last to fall. But if the walls are very steep, the sand will not form a pronounced crater and will fall more uniformly. In this case, the sand on the very top layer will be the last to fall.
That is a great find. Mostly I liked it because it confirmed what I thought, that for a standard “egg timer” type hourglass, near the bottom before the V was nearly the last to go.