8.5. What is the maximum number of members that can be in a sequence of non-zero integers, for which the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative?
Answer: 16.

Solution: Estimation. Assume there are no fewer than 17 numbers. Construct a table with 7 columns and 11 rows, in which the first 17 numbers are arranged.

| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ |
| $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ |
| $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ |
| $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ |
| $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |
| $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ |
| $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ |
| $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ |
| $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ |
| $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |

Consider the sums by rows:

$$
a_{1}+\ldots+a_{7}, \ldots, \quad a_{2}+\ldots+a_{8}, \quad a_{11}+\ldots+a_{17}
$$

All these sums are positive, so the sum of all the numbers written is positive. But if we sum by columns, we get negative sums. The obtained contradiction shows that the number of numbers in the sequence must be less than 17.

Example: 16 numbers can be: $-5,-5,13,-5,-5,-5,13,-5,-5,13,-5,-5,-5,13,-5,-5$.

Comment: A fully justified solution - 7 points. Correct estimation - 5 points, a correct example constructed - 2 points, points are summed. The problem is not solved or solved incorrectly - 0 points.