The switches and the million light bulbs

1. Light bulb n is activated by the people whose numbers divide n.
2. Final count in $1,\dots,10^6$.
3. Conclusion: the $1,4,9,16,\dots,1\,000\,000$ bulbs remain on, in total $\boxed{1000}$.

Answer: They stay on exactly

$$ 1000 $$

bulbs: those with a perfect square index.
1) Mathematical model
The $n$ light bulb is activated by the people whose numbers divide $n$.
Number of bulb actuations $n$:

$$ \tau(n)\quad (\text{number of positive divisors of }n). $$

Since it starts off and each operation inverts the state, it remains on if and only if $\tau(n)$ is odd.
2) When is $\tau(n)$ odd?
Dividers usually come in pairs:

$$ (d,\,n/d). $$

There is a divisor "without a partner" only when:

$$ d=n/d\iff d^2=n, $$

that is, when $n$ is a perfect square.
Therefore:

• if $n$ is not square, $\tau(n)$ is even;
• if $n=k^2$, $\tau(n)$ is odd.

3) Final count in $1,\dots,10^6$
The perfect squares in that range are:

$$ 1^2,2^2,\dots,1000^2, $$

because

$$ 1000^2=1\,000\,000. $$

There are exactly 1000 of them.
Conclusion: the light bulbs remain on

$$ 1,4,9,16,\dots,1\,000\,000, $$

in total

$$ \boxed{1000}. $$