Consider a 0/1 matrix whose rows correspond to students and whose columns correspond to months, where each entry is 0 or 1 according to which group that student was in on that month. The condition (every pair of students are in different groups at least once) just means that the rows are all different. If there are $ n$ students where $ 2^{k-1} < n \leq 2^k$, this can obviously be done with $ k$ months but no less, since there are only $ 2^{k-1}$ binary vectors of length $ k-1$, and on the other hand you can easily create $ 2^k$ distinct vectors by simply enumerating all of them. Therefore, the answer is 4 months when there are 16 students. The algorithm is just to enumerate the students $ 0, 1\ldots, 15$, take the 4-digit binary representation of these numbers, and placing them in groups on month $ i$ according to the value of the $ i$th digit.