I did it like this:

You want to choose egg drop levels for first egg so that ideally each worst case scenario requires the same number of total drops (impossible but goal is to get close).

let s(n) be the sum of function f(k)=n-k+1 from k=1:n.

s(n)=n^2/2+n/2

For what minimum n is s(n)>=100?

Solving for n^2+n-200>=0 gives n>=13.65, so then the minimum number of drops (and starting level) required in worst case scenario is 14. So n=14.

Starting level is f(1)=n If first egg breaks, drop 2nd at 1:n-1. If it doesn’t, drop again at f(2)=n+n-1. If it breaks, drop second at n+1:n+n-2. And so on.

So that gives 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99. This is one of several possible solutions.