Solution:

The answer is $\mathbf{( E )}$. The switch in position $n$ is touched once for each positive divisor of $n$. So the $n$-th bulb remains on at the end if and only if $n$ has an odd number of divisors. This happens only for perfect squares: indeed if $n$ is not a perfect square, we can pair up its divisors by forming all the pairs of the type $(d, n / d)$, and this tells us that they are even in number. If $n=m^{2}$ is a perfect square, we can pair up all its divisors except $m$ by again pairing $(d, m^{2} / d)$: hence $m^{2}$ has an odd number of divisors. At this point, it is easy to check that the only perfect square among the possible answers is $97^{2}=9409$.