The answer is 7 times

1. Switch two opposing switches
2. Switch two adjacent switches
3. Switch two opposing switches
4. Switch a single switch
5. Switch two opposing switches
6. Switch two adjacent switches
7. Switch two opposing switches

Now whatever the starting position was and irrelevant how you turned the table, all lights will be on.

FIRST: ASSUME we have an even amount of switches that are on and an even amount is off.
Since 0/4 switches=on means the light is on, we assume therefore there are two off and two on:

10
01

01
01

11
00

Three different possibilities (there are six, but they are equivalent in pairs, 0 and 1 mean different positions, it is irrelevant if 1=on or if 1=off). Now you switch two diagonally opposing switches. The first one will be solved - the other two mutate into:

11
00

10
10

The tables are turned - but you are still sure that two adjacent switches are on.
Now switch two adjacent switches, e.g. upper ones.
The red on is solved. The blue one mutates into:

01
10

Now you turn the table and whatever you do, switching two opposing switches will solve the problem.

So now we have the problem solved for an even amount of switches that were on in the starting position. In case we have not solved yet, means that there was an odd amount of switches on.

So switch one!

Now you are back to even and you repeat the procedure we did assuming the even amount: opposing-adjacent-opposing.

Number of floor = N
number of times you have to drop the balls = t

Let's partition the floor into 2 parts seperated by the first throw position. Part A, whatever is on top and part B whatever is under. Say you need to drop the ball k times to solve Part A, then to solve part B when the first ball breaks, you would need to drop the ball k+1 times. Which means that for any number of floor N, you would need to add k+1 floors to increase the number of throw required by 1.

This means that the number of try required is the ith triangular number which is >100 aka.

14

ok