100 Lockers and Toggled Lights

Bulb $n$ is flipped by person $k$ exactly when $k\mid n$. Every divisor of $n$ lies between $1$ and $100$, so bulb $n$ is flipped once for each divisor of $n$, that is, $d(n)$ times in total. It starts off, so it ends on if and only if $d(n)$ is odd.

Pair each divisor $k$ of $n$ with $n/k$. This is an involution on the set of divisors. Its only fixed point would be a divisor with $k=n/k$, that is, $k^2=n$. The divisors therefore split into pairs $\{k,n/k\}$ with $k\ne n/k$, plus the single divisor $\sqrt n$ when $n$ is a perfect square. Hence $d(n)$ is odd exactly when $n$ is a perfect square.

The perfect squares in $\{1,\dots,100\}$ are $1^2,2^2,\dots,10^2$. There are ten of them.

Generalisation. With $N$ bulbs and $N$ people, exactly $\lfloor\sqrt N\rfloor$ bulbs are on.