Phrasing the statement into mathematical terms, one can see that the $i$th person opens the $j$th door if and only if $i$ divides $j$; thus the problem becomes studying which numbers have an odd number of divisors.
Recall now the fundamental theorem of arithmetic, which states that each natural number $n$ can be factored as a unique product of powers of primes, that is, $$n=\prod _{i=1}^k{p}_i^{{\alpha }_i}=2^{{\alpha }_1}{3}^{{\alpha }_2}\cdots .$$ Prime numbers are the building blocks of the natural numbers, so for a number to be a divisor of $n$, it must have smaller or equal ${\alpha }_i$'s than $n$ when factorized, which therefore leaves a total of $$d(n)=\prod _{i=1}^k({\alpha }_i+1).$$ We want this product to be odd, so each individual factor must be odd itself – otherwise the entire expression becomes even –, hence each ${\alpha }_i$ is even, so $n$ is a perfect square! The converse is also true, since each perfect square has squared primes as factors. Therefore, since there are $10$ perfect squares between $1$ and $100$, namely $1$, $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$, and $100$, these positions will stay open at the end.