Problem 1802

Solution:

Answer. $n=340$.

The prime factorization of $2019$ is $3 \cdot 673$. Let $p=673$.

For each integer $k$, color the three numbers $k p-1$, $k p$, $k p+1$ red, and the six numbers $k p+\frac{p-5}{2}$, $k p+\frac{p-3}{2}$, $k p+\frac{p-1}{2}$, $k p+\frac{p+1}{2}$, $k p+\frac{p+3}{2}$, $k p+\frac{p+5}{2}$ blue. Now the integers are colored periodically. In a period of length $p=673$, there are $3$ red integers, then $332$ uncolored integers, then $6$ blue integers and finally $332$ uncolored integers.

The sum of the integers in a red interval is $3 k p = 2019 \cdot k$, and the sum of the integers in a blue interval is $6\left(k p+\frac{p}{2}\right)=2019 \cdot (2 k+1)$. So if there is a colored interval (we mean a maximal one throughout) in the given $n$ consecutive integers, one can choose it. It is easy to see, that among any $340=332+(6-1)+(3-1)+1$ consecutive integers, there must be a colored interval. Thus the smallest $n$ (that we look for) satisfies $n \leq 340$.

Now we will show that it is not possible to choose consecutive integers in the desired way from the set $A=\{335,336, \ldots, 673\}$ ($|A|=339$ and thus $n \geq 340$). Assume that there exists $\{a, a+1, \ldots, b\} \subseteq A$ such that
$$2019 \mid a+(a+1)+\cdots+b=\frac{(b-a+1)(a+b)}{2}$$
That means either $673 \mid b-a+1$, or $673 \mid a+b$. Since
$$0<1 \leq b-a+1 \leq 339<673$$
$673$ must divide $a+b$. Taking into account that
$$671=335+336 \leq a+b \leq 673+673=2 \cdot 673$$
we conclude that $a+b$ must be $673$ or $2 \cdot 673$. It means either $a=335$ and $b=338$, or $a=336$ and $b=337$, or $a=b=673$. But $2019 \nmid 335+336+337+338=1346$, $2019 \nmid 336+337=673$ and $2019 \nmid 673$, a contradiction.