A quick computer check shows that $7272$, $8484$ and $9696$ have the same (maximum) number of divisors ($24$):
{1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72, 101, 202, 303, 404, 606,
808, 909, 1212, 1818, 2424, 3636, 7272},
{1, 2, 3, 4, 6, 7, 12, 14,
21, 28, 42, 84, 101, 202, 303, 404, 606, 707, 1212, 1414, 2121,
2828, 4242, 8484},
{1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96, 101,
202, 303, 404, 606, 808, 1212, 1616, 2424, 3232, 4848, 9696}
so if I understand your question (which of these is minimum?), the answer is: $7272$.
The original question did not state that the solution must be done by hand, but in that case, @MishaLavrov's approach (below) is the best:
Recognize that $abab = ab \cdot 101$, so the factors of $abab$ are maximized when $ab$ has the most factors. It is a simple matter to find (by hand) that $72$, $84$ and $96$ have the same (maximum) number of factors.
The rest is easy.
Best Answer
Hint (as requested): Think about other bases!
Though I'll remark that I dislike these types of problems, especially in competition mathematics, since you can describe any suitable formulation and use it to justify practically any answer - as evidenced by Matti P's comment.