VI OBM

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 $(0,g)$, $(1,g+d)$, $(2,g+2d)$, $\dots$, $(16,g+16d)$, where all arithmetic is done mod 17. On the final day, take the groups as $$\begin{array}{l}(0,0),(0,1),\dots ,(0,16)\\ (1,0),(1,1),\dots ,(1,16)\\ (2,0),(2,1),\dots ,(2,16)\\ ⋮\\ (16,0),(16,1),\dots ,(16,16)\end{array}$$

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 $18\cdot 17\cdot \left(\genfrac{}{}{0ex}{}{17}{2}\right)=\left(\genfrac{}{}{0ex}{}{289}{2}\right)$ pairs available and we have just shown that each occurs at least once. Hence each occurs at most once, as required.