100 switches

1. Step 1 · Switch k is flipped by each person whose number divides k

   $\displaystyle \text{switch } 12\colon \text{ people } 1,\, 2,\, 3,\, 4,\, 6,\, 12$

   Person $m$ flips switches $m, 2m, 3m, \dots$, so a switch is flipped once for each of its divisors.

2. Step 2 · On at the end means an odd number of flips

   $\displaystyle \text{divisors pair up}\colon 2 \times 6,\, 3 \times 4 \dots$

   Divisors come in pairs that multiply to $k$. Only a perfect square has one divisor paired with itself, so only squares have an odd count.

3. Step 3 · Perfect squares up to 100

   $\displaystyle 1,\, 4,\, 9,\, 16,\, 25,\, 36,\, 49,\, 64,\, 81,\, 100 = 10$

   Count the squares of 1 to 10.

Answer$\displaystyle 10$