Thirty numbers are arranged on a circle in such a way that each number equals the absolute difference of its two neighbors. Given that the sum of the numbers is $2000$, determine the numbers.
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.