On a circle line, there are 2018 points. Each of these points is labeled with an integer. Each number is greater than the sum of the two numbers that immediately precede it in a clockwise direction.

Determine the maximum possible number of positive numbers that can be among the 2018 numbers.

(Walther Janous)
We denote the points in clockwise order as $a_{0}, a_{1}, \ldots, a_{2017}$ with cyclic notation, i.e., $a_{k+2018}=a_{k}$ for all integers $k$.

Lemma 1. In a valid arrangement, no two non-negative numbers can be adjacent.

![](https://cdn.mathpix.com/cropped/2024_04_17_a338358c8950f43c7d7dg-09.jpg?height=928&width=1265&top_left_y=119&top_left_x=447)

Figure 8: Problem 4, Solution $1 \mathrm{c}$

Proof. Suppose there are two adjacent numbers $a_{k-1}$ and $a_{k}$, both non-negative. Then it follows that $a_{k+1} > a_{k} + a_{k-1} \geq a_{k}$, where the first inequality comes from the condition in the problem statement and the second from $a_{k-1} \geq 0$. Thus, $a_{k}$ and $a_{k+1}$ are both non-negative, and analogously, $a_{k+2} > a_{k+1}$, then $a_{k+3} > a_{k+2}$, and so on, until we finally have $a_{k+2018} > a_{k+2017} > \cdots > a_{k+1} > a_{k} = a_{k+2018}$, a contradiction.

Therefore, at most every second number can be non-negative. We now show that even this is too many non-negative numbers.

Lemma 2. In a valid arrangement, not every second number can be non-negative.

Proof. Suppose this were the case, so without loss of generality, $a_{2k} \geq 0$ and $a_{2k+1} < 0$. Then $a_{3} > a_{1} + a_{2} \geq a_{1}$, where again the first inequality comes from the problem statement and the second from $a_{2} \geq 0$. Analogously, we get $a_{5} > a_{3}$, then $a_{7} > a_{5}$, and so on, until $a_{1} = a_{2019} > a_{2017} > a_{2015} > \cdots > a_{3} > a_{1}$, a contradiction.

In summary, an arrangement with more than 1009 non-negative numbers is not possible, because then, by the pigeonhole principle, there must be two non-negative numbers adjacent somewhere, which is not allowed according to Lemma 1, and an arrangement with exactly 1009 non-negative numbers contradicts either Lemma 1 or Lemma 2.

With 1008 positive and 1010 negative numbers, we find, for example, the arrangement

$$
-4035, 1, -4033, 1, -4031, 1, -4029, \ldots, 1, -2021, 1, -2019, -2017
$$

whose correctness we can easily verify.

(Note: Many other arrangements with 1008 positive numbers are possible. For example, if we decrease each negative number by 1 in a valid arrangement, we obtain another possible arrangement. Furthermore, it can be shown that every possible arrangement contains three consecutive negative numbers and otherwise alternates between positive and negative numbers, and that for any choice of positive numbers, a solution exists.)

(Birgit Vera Schmidt)