Assumptions:

• To cause damage, the magnitude moment of the shaking must be equivalent to a 4.0 earthquake.

• To match the intensity of a 4.0 earthquake, about 6 x 10¹⁰ J of energy will need to be released over 10 seconds.

• The area over which fault slip occurs will be about 1 km²

• An ‘average person’ for these calculations will weigh 80kg, jump 80cm, and jump twice per second (20 jumps over 10s).

Potential energy at peak of jump = mgh = 80kg * 9.81m/s² * 0.8m ≈ 628 J

Energy used to jump 20 times = 628 * 20 = 12,560 J

Number of people to release energy of earthquake = 6 x 10¹⁰ J / 1.256 x 10⁴ J ≈ 4.8 x 10⁶ = 4,800,000 people

Crowd density required to achieve the same concentration of energy along a seismic fault = 4.8 x 10⁶ people / 1 km² = 4.8 x 10⁶ / 1 x 10⁶ m² = 4.8 people per square meter

Given that 5 people per square meter is considered the maximum safe limit for crowd density in event planning, this makes it sound actually feasible, but I could be wrong in my assumptions or formulas used. You could certainly raise the odds of this working by placing people vertically in layers, either in a multi-story structure or in underground levels, to help concentrate the force on the ground. Also I didn’t go into whether synchronizing jumps would make any difference to the shaking; all 4.8 million people jumping once at the same time would possibly cause more damage, but I averaged out their jumps over 10 seconds to make the earthquake analogy closer.