4. A circle is inscribed with 1001 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 4, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers.
Answer: 3009.

Solution. Due to the odd number of total numbers, there will be three consecutive numbers $x, y$, and $z$ such that $x>y>z$. Adding the inequalities $y-z \geqslant 4$ and $y+z \geqslant 6$, we get $y \geqslant 5$. Then $x \geqslant y+4 \geqslant 9$. A number not less than 9 has been found. The remaining numbers can be divided into 500 pairs of adjacent numbers. Therefore, the sum of all numbers $S \geqslant 9+500 \cdot 6=3009$.

The lower bound for $S$ obtained can be achieved if one of the numbers is 9, and then, in a circle, alternating numbers 5 and 1 are placed.

Evaluation. 14 points for a complete solution. 5 points for an example without estimation, 7 points for estimation without an example.