Problem 1332

1. Let's denote the 30 numbers as $a_1, a_2, \ldots, a_{30}$ arranged in a circle. This means $a_{31} = a_1$.
2. According to the problem, each number equals the absolute difference of its two neighbors. Therefore, we have:
$$a_i = |a_{i-1} - a_{i+1}|$$
for $i = 1, 2, \ldots, 30$.
3. Given that the sum of the numbers is 2000:
$$\sum_{i=1}^{30} a_i = 2000$$
4. Let's consider the solution provided: the numbers are composed of 10 blocks of three numbers, and each block has the numbers $100, 0, 100$ in that order.
5. We can verify this solution by checking if it satisfies the conditions:
- Each block is $100, 0, 100$.
- The absolute difference condition:
$$|100 - 0| = 100, \quad |0 - 100| = 100, \quad |100 - 100| = 0$$
This satisfies the condition for each block.
6. Since there are 10 such blocks, the total sum is:
$$10 \times (100 + 0 + 100) = 10 \times 200 = 2000$$
This matches the given sum of 2000.
7. Therefore, the numbers can indeed be arranged in 10 blocks of $100, 0, 100$.

The final answer is $\boxed{ 100, 0, 100 }$ repeated 10 times.