Answer. Considering the first number mentioned in the condition is 6, all numbers with odd indices are equal to 6, and all numbers with even indices are equal to -4. Solution. Let's denote the numbers in order as $a_{1}, a_{2}, \ldots, a_{100}$. Adding all 100 inequalities $a_{i}+a_{i+1}+\ldots+a_{i+5} \leq 6, i=1,2, \ldots, 100, \quad$ we get $6 \cdot 100=6\left(a_{1}+a_{2}+\ldots+a_{100}\right) \leq 100 \cdot 6=600$. From this, it follows that all inequalities are, in fact, equalities. From the equalities $a_{i}+a_{i+1}+\ldots+a_{i+5}=6=a_{i+1}+a_{i+1}+\ldots+a_{i+6} \quad$ it follows that $a_{i}=a_{i+6}, i=1, . ., 100$, meaning that all numbers with odd indices are equal to each other, and all numbers with even indices are equal to each other. We can assume that all numbers with odd indices are equal to 6, then their sum is 300, which means the sum of all numbers with even indices is -200. Therefore, all numbers with even indices are equal to -4.
9.5. One hundred numbers are written in a circle with a total sum of 100. The sum of any 6 consecutive numbers does not exceed 6. One of the written numbers is 6.

Find the other numbers.