Each day student $X$ is with 16 different students. But there are only 288 students available, so it can be done for at most $288/16 = 18$ days.

We now show explicitly how it can be done for 18 days. Label the students as $(m, n)$, where $0 \le m, n \le 16$. Now for $0 \le d, g \le 16$, take group $g$ on day $d$ to be $(0, g)$, $(1, g+d)$, $(2, g+2d)$, $\ldots$, $(16, g+16d)$, where all arithmetic is done mod 17. On the final day, take the groups as
$$\begin{array}{l} (0, 0), (0, 1), \dots, (0, 16) \\ (1, 0), (1, 1), \dots, (1, 16) \\ (2, 0), (2, 1), \dots, (2, 16) \\ \vdots \\ (16, 0), (16, 1), \dots, (16, 16) \end{array}$$

Note first that for any given day the groups include all the students. That is obvious for the last day. Suppose we want to find the group for student $(a, b)$ on day $d$. Take $g = b - ad$ mod 17, then $(a, b) = (a, g + ad)$, so $(a, b)$ is in group $g$. Since the groups include just 289 students and each student is included at least once, it follows that each student is included exactly once.

Now consider the pair $(a, b), (A, B)$. If $a = A$, then they are together on the last day. So suppose $a \ne A$. Take $d = (b - B)(a - A)^{-1}$ mod 17 and take $g = b - ad$ mod 17. Then $(a, b) = (a, g + da)$ and $(A, B) = (A, g + dA)$, so they are together in group $g$ on day $d$. But there are only $18 \cdot 17 \cdot \binom{17}{2} = \binom{289}{2}$ pairs available and we have just shown that each occurs at least once. Hence each occurs at most once, as required.