Is it possible to place 2019 natural numbers along a circumference so that, for any two of these numbers, the proportion of the largest to the least is a prime number?
Ok, what I am clear about this problem is that for the ratio to be a number first, it means that given any two numbers, its mcd can not be 1, because in this way we make sure that at least the proportion is a natural number ... However, I am not very clear on how to interpret the part in which it says that they are located along a circumference, that is, How different is it if we place it on a straight?
Let us denote the numbers by $x_0,\ldots,x_{m-1}$, where in your case $m = 2019$. Let $P(x)$ denote the number of prime factors of $x$. Your condition implies that $P(x_i) = P(x_{i+1}) \pm 1$, and in particular $P(x_i) \equiv P(x_{i+1}) + 1 \pmod{2}$, where the indices are modulo $m$. Summing this over all $i$, we get the contradictory $0 \equiv 1 \pmod{2}$, since $m$ is odd.
In contrast, if $m$ is even, then there is always a solution. I will illustrate this with $m = 6$: $$ 1,2,4,8,4,2. $$ The general solution is similar.