1064. Several natural numbers, more than three, are written in a circle, the sum of which is 19. The sums of any three consecutive numbers are equal to each other. What numbers were written in the circle?
$\triangle$ Let's denote these numbers in sequential order as $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$, where $n-$ is the total number of numbers. Then

$$
a_{1}+a_{2}+a_{3}=a_{2}+a_{3}+a_{4}
$$

From this, $a_{1}=a_{4}$.

Similarly, from the equalities

$$
a_{2}+a_{3}+a_{4}=a_{3}+a_{4}+a_{5}, \quad a_{3}+a_{4}+a_{5}=a_{4}+a_{5}+a_{6}
$$

it follows that $a_{2}=a_{5}$ and $a_{3}=a_{6}$. And so on.

Let

$$
a_{1}=a, \quad a_{2}=b, \quad a_{3}=c
$$

Consider three cases.

1) Suppose the last of the $n$ numbers is $-a: a_{n}=a$.

Then the numbers are arranged in the following order:

$$
a, b, c, a, b, c, \ldots, a, b, c, a
$$

By the condition, we get:

$$
a+b+c=c+a+a=a+a+b
$$

since the last number stands next to the first on the circle.

From this, $a=b=c$.

2) Suppose $a_{n}=b$. Then the numbers are arranged as follows:

$$
a, b, c, a, b, c, \ldots, a, b, c, a, b
$$

In the same way, we establish that $a=b=c$.

3) Suppose $a_{n}=c$. This case is impossible, since then the sum of all numbers would be divisible by 3.

It turns out that all the numbers on the circle are equal to each other. Since the number 19 is prime, this is possible only when each of the numbers is equal to 1, and $n=19$.

Answer: 19 numbers, each of which is equal to 1.