This problem is actually quite easy. Suppose for the sake of contradiction that no blue label is a multiple of 3.Focus on the point labels modulo 3. Then, any two consecutive labels must be different. Assume now that we have a chord with endpoints $x,y\in\{0,1\}$. Delete all points labeled $2$ on that chord. In this new labeling we have, modulo 2, $x+y$ $0-1$ edges. But note that by adding back the $2$s, an edge $0-1$ with one $2$ in-between would give one $0-2$ edge and one $1-2$ edge. Meanwhile, an edge $0-0$ with one $2$ in-between would give two $0-2$ edges and zero $1-2$ edges, and one $1-1$ edge would give zero $0-2$ edges and two $1-2$ edges. From here it is clear that on each chord we have $$\text{nr. of 0-2 edges}+\text{nr. of 0-1 edges}\equiv \text{nr. of 1-2 edges}+\text{nr.of 0-1 edges}\equiv x+y(\text{mod} 2).$$Summing all over the chords we obtain $$\text{nr. of 0-2 edges}+\text{nr. of 0-1 edges}\equiv \text{nr. of 1-2 edges}+\text{nr.of 0-1 edges}\equiv 2019\equiv 1(\text{mod} 2)$$meaning that $X_0\equiv X_2\not\equiv X_1(\text{mod} 2)$, where $X_i$ is the number of numbers from $0,1,...,N$ that are $i$ modulo 3. It's easy to see that this is not possible.

Suppose no blue label is a multiple of 3. Interpret the point labels modulo 3. Let $e_{01}$ denote the number of 0-1 edges; define $e_{02}$ and $e_{12}$ similarly. Note that apart from the 4038 chord endpoints, each marked point has even degree. Thus (recall 2019 is odd) \begin{align*} \sum_{v \; \text{labeled} \; 0} \deg v & = e_{01} + e_{02} \equiv 1 \pmod{2}\\ \sum_{v \; \text{labeled} \; 1} \deg v & = e_{01} + e_{12} \equiv 1 \pmod{2}\\ \sum_{v \; \text{labeled} \; 2} \deg v & = e_{02} + e_{12} \equiv 0 \pmod{2}. \end{align*}Solving the above system we get \[(e_{12}, e_{01}, e_{02}) = (\text{odd}, \text{even}, \text{odd}) \; \text{or} \; (\text{even}, \text{odd}, \text{even}).\]However, by the hypothesis, it follows that \[(e_{12}, e_{01}, e_{02}) \in \big\{(a, a, a), (a+1, a, a), (a+1, a+1, a)\big\}\]for some $a$, contradiction!

Consider all labels mod $3$. Let $S$ be the set of intersection points of chords and let $S_i$ be the subset of points of $S$ labeled $i$. Let $d(s)$ be the number of chords containing $s$ for $s \in S$. Assume for the sake of contradiction that no two points in a segment have the same label. Notice that there are $\left\lfloor\frac{N + 3}{3}\right\rfloor$ segments labeled $0$, $\left\lfloor\frac{N + 2}{3}\right\rfloor$ labeled $1$ and $\left\lfloor\frac{N + 1}{3}\right\rfloor$ labeled $2$. Now notice that $$n + 2 \sum_{s \in S_0} d(s) = \left\lfloor\frac{N + 1}{3}\right\rfloor + \left\lfloor\frac{N + 2}{3}\right\rfloor$$ As both count the number of segments labeled $1$ or $2$. Similarly we get $$n + 2\sum_{s \in S_1} d(s) = \left\lfloor\frac{N + 1}{3}\right\rfloor + \left\lfloor\frac{N + 3}{3}\right\rfloor$$ And $$2\sum_{s \in S_2} d(s) = \left\lfloor\frac{N + 3}{3}\right\rfloor + \left\lfloor\frac{N + 2}{3}\right\rfloor$$ Adding up gives $$2n + 2\sum_{s \in S} d(s) = 2\left(\left\lfloor\frac{N + 3}{3}\right\rfloor + \left\lfloor\frac{N + 2}{3}\right\rfloor + \left\lfloor\frac{N + 1}{3}\right\rfloor\right)$$ Now notice that $2(N + 1) = 2n + 2\sum_{s \in S} d(s)$ since both count the number of pairs $(\text{point}, \text{segment})$ where $\text{point}$ is an endpoint of $\text{segment}$. We then get $$N + 1 = \left\lfloor\frac{N + 1}{3}\right\rfloor + \left\lfloor\frac{N + 2}{3}\right\rfloor + \left\lfloor\frac{N + 3}{3}\right\rfloor$$ Which is only possible if $\left\lfloor\frac{N + 3}{3}\right\rfloor = \left\lfloor\frac{N + 2}{3}\right\rfloor = \left\lfloor\frac{N + 1}{3}\right\rfloor = \frac{N + 1}{3}$. However this implies that all three individual sums are equal which implies $n$ even by taking mod $2$, a contradiction.

This problem now makes my top-10 list of most bait problems I have ever seen. If $2019$ is replaced by $n$, does there even exist any arrangement satisfying the conditions, for any $n$? I was unable to find a single one.

v_Enhance wrote: If $2019$ is replaced by $n$, does there even exist any arrangement satisfying the conditions, for any $n$? I was unable to find a single one.

On a related note, would the problem be false with an even number of chords? I tried to construct configurations with 4 or 6 chords that contradicted it without much success.

Even though the problem is placed a little high, the idea involved is truly beautiful! Here's my solution (Nothing new): FTSOC assume that no two consecutive marked points are same modulo $3$. Replace each label by either $0$ or $\pm 1$, depending on its remainder modulo $3$. The main observation is that the total number of neighbors of each marked point of category $\text{(ii)}$ is even (since each edge which enters must also exit). Let $a$ be the number of unordered pairs of marked points having labels in $\{0,1\}$; similarly define $b$ for $\{0,-1\}$ and $c$ for $\{1,-1\}$. Then $a+b$ is the total number of marked points having $0$ as their label. Using our observation, and since there are an odd number of marked points of category $\text{(i)}$ with label $0$, so we get that $a+b$ is odd. Through similar reasoning one gets that $b+c$ is even, and $c+a$ is odd. This means that $a$ has different parity than both $b,c$. Now, by the given condition, one finds that $$a=\left\lfloor\frac{N+2}{3}\right\rfloor,b=\left\lfloor\frac{N+1}{3}\right\rfloor ,c=\left\lfloor\frac{N}{3}\right\rfloor +1$$So we must have either $a=b$ or $a=c$, and thus we arrive at the desired contradiction. Hence, done. $\blacksquare$

The problem seemed to be hard. Perfect Solvers of the problem : 6 participants/196 participants. among them, there is an absolute winner, perfect score ! Marvelous !

CantonMathGuy wrote: However, by the hypothesis, it follows that \[(e_{12}, e_{01}, e_{02}) \in \big\{(a, a, a), (a+1, a, a), (a+1, a+1, a)\big\}\]for some $a$, contradiction! juckter wrote: Notice that there are $\left\lfloor\frac{N + 3}{3}\right\rfloor$ vertices labeled $0$, $\left\lfloor\frac{N + 2}{3}\right\rfloor$ labeled $1$ and $\left\lfloor\frac{N + 1}{3}\right\rfloor$ labeled $2$. Can everyone help me explain it? I don't understand Sorry

This is counting the number of positive integers not exceeding $N$ that are 0,1,2 mod 3, respectively.

rmtf1111 wrote: On a circle, Alina draws $2019$ chords, the endpoints of which are all different. A point is considered marked if it is either $\text{(i)}$ one of the $4038$ endpoints of a chord; or $\text{(ii)}$ an intersection point of at least two chords. Alina labels each marked point. Of the $4038$ points meeting criterion $\text{(i)}$, Alina labels $2019$ points with a $0$ and the other $2019$ points with a $1$. She labels each point meeting criterion $\text{(ii)}$ with an arbitrary integer (not necessarily positive). Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference. Alina finds that the $N + 1$ yellow labels take each value $0, 1, . . . , N$ exactly once. Show that at least one blue label is a multiple of $3$. (A chord is a line segment joining two different points on a circle.) This is also surprisingly easy for EGMO despite being a P6! We consider the whole system of numbers modulo $3$ so Alina only writes the numbers $0, 1, 2$ and let us say that no blue label is multiple of $3$. We see that if yellow label between points numbered $i$ and $j$ is represented by $s_{ij}$, then $f : (i, j) \rightarrow s_{ij}$ is a bijection which is the key here. If $a_{01}, a_{02}$ and $a_{12}$ (no $1-1$ or $0-0$ or $2-2$ labels exist due to the assumption) represent number of $0-1$, $0-2$ and $1-2$ edges respectively, then the sum of segment degree of all such vertices numbered $0$ is $1 \pmod{2}$ since every number numbered $0$ has either odd or even degree the previous possible if and only if the vertex is on the circumference of which there are $2019$ instances and similar condition holds true for sum of segment degree of all vertices numbered $1$. However, sum of segment degrees of all vertices numbered $2$ is $0 \pmod{2}$ because every vertex numbered $2$ has even number of chords through it. However, the above results also translate to $a_{01} + a_{02} \equiv 1 \pmod{2}, a_{01} + a_{12} \equiv 1 \pmod{2}$ and $a_{12} + a_{02} \equiv \equiv a_{01} + 1 \equiv 0 \pmod{2}$ which implies that $a_{12} \equiv a_{02} \pmod{2}$. We see that due to the condition the values $1, 2, 3 \dots N$ are taken uniquely by labels then $\lvert a_{01} - a_{02} \rvert, \lvert a_{02} - a_{12} \rvert, \lvert a_{01} - a_{12} \rvert \leq 1$ and therefore, we get that $a_{12} \equiv a_{02} \equiv a_{01} \pmod{2}$ or $a_{12} \not\equiv a_{02} \pmod{2}$ which is the desired contradiction.

Suppose for the sake of contradiction that no blue labels are multiples of $3$. Work entirely modulo $3$. Every segment is one of "$\{0,1\},\{0,2\},\{1,2\}$". We count the number of segments each with $0$, $1$, or $2$ as an endpoint. Note that any type (ii) intersection contributes to an even number of these, hence the number of segments with $0$ as an endpoint is odd, as is the number of segments with $1$ as an endpoint, but the number of segments with $2$ as an endpoint is even. Therefore, the number of segments of the form $\{0,1\}$ is opposite parity from the number of segments of the form $\{0,2\}$ and $\{1,2\}$, hence the number of segments which have yellow label $1$ is opposite parity from the number of segments which have yellow label $0$ or $2$. But because the yellow labels are $0,\ldots,N$ (each with multiplicity $1$) this is evidently impossible. $\blacksquare$