[Bring_Back_Shantz]

Spoiler!

I managed to do better and get it down to 15 minimum drops.
First drop is N=1
Second drop is N =2
First floor you drop it from is floor F

Best case is F = 14
For each drop you go up F-(N-1) floors (assuming it doesn't break).
Drop 1 = 14-(1-1) = floor 14
Drop 2 = 14 + (14-(2-1)) = floor 27
Drop 3 = 27 + (14-(3-1)) = floor 39
Once it breaks you then go back to and drop it from each floor one by one from the last floor it broke on until it breaks.
So if it breaks on the 2nd drop (floor 27), you then go back to 15 and drop 1 by 1 (15-26 = 12 drops + the you already dropped)

When the first egg drops you'll have dropped it N times, and the number of drops between the safe floor, and broken floor is N-2), so you no matter where it breaks your maximum number of drops is
N + F -(N-2) = F

The minimum is where your number of big drops, and the final number of drops between that and 100 (assuming you have to go through the final interval) is a minimum.
I haven't taken the time to generalize the numbers, but it looks like it's at 14.

I'll see if I can figure out the general equation.

Edit: Okay I've generalized it
You need a starting floor where you can get to 100 before the interval between drops gets to zero otherwise you'll get to your number F and will require more steps to check up to 100, so your total number of drops will be at least F+1.
Highest floor you can get to is
F + (F-1) + (F-2) etc
it's a sum of the numbers from 1 to F
So the max you can get to is (F+1)*(F/2)
The first number that gets above 100 is 14: (14+1)*(14/2) = 105
So for any building up to 105 floors, you'll always be able to figure out how high the egg can be dropped within 14 drops.