11.4. In a row, $n$ integers are written such that the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative. For what largest $n$ is this possible?
Answer: 16

Solution: Let's provide an example for $n=16$:

$80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80$. We will prove that for $n \geq 17$, it will not be possible to write down a sequence of numbers that satisfy the conditions of the problem. Let's construct a table for the first 17 numbers in this sequence:

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

According to the condition: the sum of the numbers in each row must be negative, and in each column positive. Therefore, the sum of all the numbers in the table should be positive on one side and negative on the other. This is a contradiction.

Criteria: Only the correct answer - 1 point. Correct answer with an example - 2 points.