9.1. Over 100 days, each of the six friends visited the pool exactly 75 times, no more than once a day. Let $n$ be the number of days when the pool was visited by at least five of them. Determine the maximum and minimum possible values of the number $n$.
Answer: 90 and 25, respectively.

Solution. The total number of visits to the swimming pool by six friends over 100 days is $75 \cdot 6 = 450$, so the number of days when the pool was visited by at least five of them is no more than $\frac{450}{5} = 90$. On the other hand, if the number of such days is $n$, then the total number of visits by the friends to the pool, which is 450, does not exceed $6n + 4(100 - n) = 400 + 2n$, from which $n \geq 25$.

It remains to provide examples where these values are achieved. For this, we will write their surnames in order: first, second, ..., sixth, and so on 45 times in a row. Then we will divide them into 90 consecutive fives, these fives will visit the pool on the first 90 days. In the remaining 10 days, in this example, no one will go to the pool.

In the second case, in the first 25 days, the first 25 consecutive sixes of friends will go to the pool, and in the remaining 75 days, the remaining 75 consecutive fours will go.

Grading criteria. ($\cdot$) Estimates $n \geq 25$ and $n \leq 90$ - 2 points each, ($\cdot$) examples for $n=25$ and $n=90$ - 1 point for each. ($\cdot$) If everything is done: 7 points.