4. A circle is inscribed with 2019 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 2, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers.
Answer: 6060.

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 $y-z \geqslant 2$ and $y+z \geqslant 6$, we get $y \geqslant 4$. Then $x \geqslant y+2 \geqslant 6$. A number not less than 6 has been found. The remaining numbers can be divided into 1009 pairs of adjacent numbers. Therefore, the sum of all numbers $S \geqslant 6+1009 \cdot 6=6060$.

The lower bound for $S$ obtained can be achieved if one of the numbers is 6, and then, in a circle, the numbers 4 and 2 alternate.

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