Austrian Mathematical Olympiad, National Competition (Final Round, part 2)

Let the points be labeled $a_0, a_1, \dots, a_{2017}$ clockwise with cyclical notation, i.e., $a_{k+2018} = a_k$ for all integers $k$.

Lemma. In a valid configuration, no two neighbouring numbers can be both non-negative.

Proof. Assume that there exist neighbouring numbers $a_{k-1}$ and $a_k$ which are both non-negative. We get $a_{k+1} > a_k + a_{k-1} \ge a_k$, with the first inequality following from the problem statement and the second from $a_{k-1} \ge 0$. Since now also $a_k$ and $a_{k+1}$ are both non-negative, we analogously get $a_{k+2} > a_{k+1}$, then $a_{k+3} > a_{k+2}$, and so on, until we have $a_{k+2018} > a_{k+2017} > \dots > a_{k+1} > a_k = a_{k+2018}$, a contradiction. ■

Therefore at most every second number can be non-negative. Next we will show that these are still too many non-negative numbers.

Lemma. In a valid configuration, it is not possible that every second number is non-negative.

Proof. Assume that this is the case, so w.l.o.g. $a_{2k} \ge 0$ and $a_{2k+1} < 0$ for all integers $k$. Then we get $a_3 > a_2 + a_1 \ge a_1$, where again the first inequality follows from the problem statement and the second from $a_2 \ge 0$. Analogously we get $a_5 > a_3$, then $a_7 > a_5$, etcetera, until we have $a_1 = a_{2019} > a_{2017} > a_{2015} > \dots > a_3 > a_1$, a contradiction. ■

We can therefore summarize: A configuration with more than 1009 non-negative numbers is not possible because otherwise by the pigeonhole principle we would have two neighbouring non-negative numbers, which is not allowed according to the first lemma. A configuration with exactly 1009 non-negative numbers contradicts either the first or the second lemma.

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

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

which we can easily check for correctness.