While it is true that, at least over time, Earth will equilibrate so the mountain sinks down, displacing the denser material below, this would also mean that the material just beside the mountain is less dense, and thus there is a similar reduction in g due to this when standing beside it.
And while the increased mass is quite small relative to Earth, you are also a lot closer to it. So you need to consider both factors.
If we do a first order approximation for the change in gravity due to the height of the mountain, we get; $$ dg=\frac{2 G M h}{R^3} $$ where $R$ is the radius of Earth, $M$ is the mass of Earth and $h$ is the height of the mountain.
If we treat the mountain as a sphere with a diameter of $h$ and a density of $\rho$, then the acceleration due to gravity when standing on top of it is: $$ a=\frac{G \rho (4/3) \pi (h/2)^3}{(h/2)^2} =\frac{2 G \pi \rho h}{3} $$
And with that we end up with a ratio of the 2 forces dependent upon the density of the mountain (as both are linearly dependent upon the height) $$ \frac{a}{dg}=\frac{\pi \rho R^3}{3M} $$ So for a spherical mountain you need a density of roughly 22000 kg/m^3, e.g. 15
a mountain of osmium, to have the gravity due to the mountain increase your weight more than the extra height decreases it.
Having mass in addition to that sphere (without taking away from that sphere) would reduce that density requirement. But even if you modelled it as a cylinder where h=d to calculate the mass, and overestimated the force due to gravity by treating it as a sphere/point mass, you end up needing a density of roughly 15000 kg/m^3.
However the effect is much greater than the 0.2% suggested by userLTK. With a specific gravity of 2.7, the effect of the mass of the spherical mountain is 12% that of the effect of the increased distance from Earth. 921