11.9. In a class, there are $m$ students. During September, each of them went to the swimming pool several times; no one went twice on the same day. On October 1st, it turned out that all the numbers of swimming pool visits by the students were different. Moreover, for any two of them, there was definitely a day when the first one was at the pool and the second one was not, and a day when, conversely, the second one was at the pool and the first one was not. Find the maximum possible value of $m$. (September has 30 days.)
Answer. $m=28$.

Solution. For each natural $n$, let $X_{n}=$ $=\{1,2, \ldots, n\}$. To each student, we associate the set of all days when he went to the pool (this will be a subset of $X_{30}$). Thus, we have obtained a set of $m$ (according to the condition, non-empty) subsets of $X_{30}$. The condition is equivalent to the fact that all subsets have different numbers of elements, and none of them is contained in another; we will call such a set of subsets $x o r o-$ good. Thus, we need to find the maximum number of sets in a good family of subsets of $X_{30}$.

First, let's prove that such a set cannot contain more than 28 sets. This is obvious if the set contains a 30-element subset, since it contains any other. Therefore, we can assume that the sets in the set can consist only of $1,2, \ldots, 29$ elements (and there are no more than 29 of them). Suppose that in a good set there is a 29-element set $A$ and a 1-element set $B$. Since $B$ is not contained in $A$, they do not intersect. Then any other subset in $X_{30}$ either contains $B$ or is contained in $A$. Therefore, in this case, the good set consists only of two subsets. Finally, if the set does not contain a 1 or 29-element subset, then it already contains no more than 28 sets, which is what we needed.

It remains to provide an example of a good set of 28 subsets in $X_{30}$. For this, we will show by induction on $k \geqslant 2$ that there exists a good set $A_{1}, A_{2}, \ldots, A_{2 k-2}$ of subsets in $X_{2 k}$, and $A_{i}$ contains $i+1$ elements. In the base case $k=2$, the subsets $A_{1}=\{1,2\}$ and $A_{2}=\{1,3,4\}$ work.

Suppose that for some $k$ a required good set $B_{1}, \ldots, B_{2 k-2}$ of subsets in $X_{2 k}$ has already been constructed. Then the required good set of subsets in $X_{2 k+2}$ can be constructed as follows. Let $A_{i+1}=B_{i} \cup\{2 k+2\}$ for $i=1,2, \ldots, 2 k-2$; these sets contain $3,4, \ldots, 2 k$ elements, respectively. Finally, let $A_{1}=\{2 k+1,2 k+2\}$ and $A_{2 k}=\{1,2, \ldots, 2 k+1\}$. It is not difficult to check that they form the required good set. Thus, the induction step is proved.

Remark. The arguments in the second paragraph of the solution show that in a good set of subsets in $X_{n}$ there are no more than $n-2$ sets if $n \geqslant 4$. On the other hand, acting similarly to the second half of the solution, one can construct a good set of $2 k-1$ subsets in $X_{2 k+1}$ for $k \geqslant 1$; the base of induction is provided by the subset $A_{1}=\{1,2\}$.

One can also arrange a direct induction transition that allows constructing a good set of $n-1$ subsets in $X_{n+1}$ from a good set of $n-2$ subsets in $X_{n}$ (for $n \geqslant 5$). For such a transition, the following consideration is useful: if you take the complements of all sets of a good set, you will also get a good set.

Comment. Only the answer - 0 points.

Proved that the number of students in the class does not exceed $29-$ 0 points.

Provided an example with no more than 27 students - 0 points.

Proved only that the number of students in the class does not exceed $28-2$ points

Only provided an example with 28 students - 4 points.