66. Given natural numbers $n \geqslant k$. In a certain school, students attend $n$ clubs, and for any $k$ clubs, each student attends at least one of them, but for no $k-1$ clubs is this true. What is the smallest possible number of students in such a school?
Solution: for any $k-1$ circles, there is a student who does not attend them. Moreover, these students for different sets of $k-1$ circles are distinct. Then we get that the total number of students is no less than $C_{n}^{k-1}$. If there are exactly that many students, there is an example: for

![](https://cdn.mathpix.com/cropped/2024_06_04_defe0947e31c20461dc9g-29.jpg?height=983&width=1014&top_left_y=1710&top_left_x=521)

Fig. 20: To problem 63.
each set of $k-1$ circles, there is a unique student who does not attend them. 67. Let $n$ be a natural number. For which natural $k \leqslant n$ can numbers be placed in an $n \times n$ square such that the sum of all numbers in the table is positive, and the sum of numbers in each $k \times k$ square is negative?

Answer: for $k$ not dividing $n$. Solution. Note that if $n$ is divisible by $k$, then the $n \times n$ square can be completely divided into $k \times k$ squares, making the required arrangement of numbers impossible. Now let $n$ not be divisible by $k$. Number the rows and columns from 1 to $n$. Place the number $1-k^{2}$ in the cells where both coordinates are divisible by $k$, and place -1 in the other cells. Then the sum of the numbers in the entire table will be positive, and the sum of the numbers in each $k \times k$ square will be 0. Decrease all the numbers in the table by a small $a>0$ such that the sum of all numbers remains positive. Clearly, we will obtain the required arrangement.