Belarusian Mathematical Olympiad

Answer: 24 days. Note that from four hobby groups one can form exactly $4\cdot 3/2=6$ different pairs. Since we have exactly six pupils, for each pair of the hobby group there exists exactly one pupil attending just these two hobby groups. By condition, each hobby group is attended by either one or two pupils and there are exactly four hobby group, so in each of these days there are exactly two hobby groups attended by two pupils and there are exactly two hobby groups attended by one pupil.

Consider arbitrary day. Let this day hobby group $A$ and $B$ be attended by two pupils. Since, as was shown above, there is a pupil attending just these two hobby groups, without loss of generality we can assume that this pupil has attended hobby group $A$. Each of two remained hobby groups $C$ and $D$ has been attended by one pupil. Again, there exists a pupil attending just these hobby groups $C$ and $D$. Suppose that this pupil has attended hobby group $C$. Show that this information is sufficient to uniquely indicate for each hobby group who of the pupils has attended this hobby group.

Since in the considered day hobby group $C$ has been attended by exactly one pupil (this pupil attends hobby groups $C$ and $D$), the pupil attending $A$ and $C$ has attended hobby group $A$, and the pupil attending $B$ and $C$ has attended hobby group $B$. Further, since hobby group $A$ has been attended by two pupils (one of them attends hobby groups $A$ and $B$, and the other one attends hobby group $A$ and $C$), the pupil attending $A$ and $D$ has attended hobby group $D$. Finally, the pupil attending hobby groups $B$ and $D$ has attended hobby group $B$, since hobby group $D$ has been attended by exactly one pupil attending hobby groups $A$ and $D$.

We have exactly $4\cdot 3/2=6$ ways to choose the pair $(A,B)$ that have been attended by two pupils, and for each such pair we have two ways to fix the hobby group has been attended by the pupil attending just these two hobby groups. For hobby groups $C$ and $D$ we have two ways to choose the hobby group that has been attended by the pupil attending these hobby groups $C$ and $D$. Thus, we obtain $6\times 2\times 2=24$ different days in total.