I believe the apparent paradox is that the 'net force' is not the same thing as pressure; it's basically the vector sum of pressure. Pressure in the Earth is an isotropic distribution of stresses, so that in a north-east-up coordinate system, the upward-pushing stress is matched by the downward-pushing stress, the eastward-pushing stress is matched by the westward-pushing stress, etc. So while these sum to zero in some sense, their magnitude is the density of the rock between the point and the surface times the depth times gravity. This is like going deeper in the ocean-the forces are equal in all directions, meaning that they sum to zero in some sense and there is no directional flattening or translation, but they still increase with depth.
> I believe the apparent paradox is that the 'net force' is not the same thing as pressure
I don't think that's what OP means. He's saying that the gravitational force that acts on a given test particle that's located inside Earth at a radius r=R from the center is determined only by the amount of mass "below it", i.e. the mass inside the ball of radius R, not the matter making up the spherical shell R < r < R_E, where R_E is Earth's radius. Put differently, that (hollow) spherical shell does not cause any gravitational force on test particles inside it.
Meanwhile, you are talking about a different thing entirely: The pressure. The matter above (as well as below and next to) our test particle will obviously also experience gravity and get pulled down. So our test particle will experience an isotropic force from all sides, which is generally quantified as pressure (force per area). This pressure is obviously not zero (but no one ever claimed that) and does depend on the radius of the outer spherical shell above it, in the same way as the pressure under water depends on how deep you are.
