A Drunken Prison Guard.

Well, cell $n$ is being opened and closed as many times as it has divisors. Thus, it will only end up open if the number of divisors is odd. Therefore, only the perfect square numbered cells will be open in the end.

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A quick explanation on why perfect squares are the only numbers with an odd number of divisors; for every $d\mid n$ there exists a $\frac{n}{d}\mid n$, so divisors come in pairs. The only non-pair that can occur is when $d=\frac nd$, or when $n=d^2$. Then $d$ is the only divisor not part of a pair, and thus makes the total number of divisors odd.

Under @shoover's interpretation of the problem where the circularity is relevant, here's a simple python script that prints the open cells after each pass:

doors = [False for n in range(0, 100)]

def print_f():
    print [i + 1 for (i, bool) in enumerate(doors) if bool]
        # code is 0-based, question statement is 1-based

paces = 0
step = 1

while step < 100:
    location = paces % 100
    doors[location] = not doors[location]
    if location + step >= 100:
        step += 1
        print_f()
    paces += step

The final result is

[1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 20, 21, 22, 24, 27, 33, 34, 35, 36, 38, 39, 40, 41, 44, 47, 49, 50, 51, 53, 57, 58, 62, 64, 65, 71, 73, 76, 77, 79, 80, 81, 86, 87, 88, 90, 92, 94, 96, 99]

which doesn't look like it has an elegant description to me. My guess would therefore be that the conventional version of the problem was the intended one.