I think you can do this with a big M approach using MIDCP capability which you have with MOSEK as solver.

Let M be the smallest upper bound you can impose on x. Given that sum(x) == 1, M can be set to 1. Let c be some very small constant. I donâ€™t know whether c equal to zero will work given solver tolerance. There has to be some slop somewhere, perhaps the slop in the solver is good enough.

cvx_begin
variable x(100,1)
variable y(100,1) binary
x <= c + M*(1-y)
sum(y) >= 50
sum(x) == 1; % NOTE: this must have == , not =
x>=0;
maximize(<some quadratic objective>)
cvx_end