Somebody placed digits $1,2,3, \ldots , 9$ around the circumference of a circle in an arbitrary order. Reading clockwise three consecutive digits you get a $3$-digit whole number. There are nine such $3$-digit numbers altogether. Find their sum.
1. **Understanding the problem**: We need to find the sum of all possible 3-digit numbers formed by reading three consecutive digits from a circle containing the digits \(1, 2, 3, \ldots, 9\) in any order.

2. **Observation**: Each digit from \(1\) to \(9\) will appear exactly once in each of the hundreds, tens, and units places across all the 3-digit numbers formed.

3. **Summing the contributions**:
    - Each digit \(d\) (where \(d\) ranges from \(1\) to \(9\)) will contribute \(100d\) when it is in the hundreds place, \(10d\) when it is in the tens place, and \(d\) when it is in the units place.
    - Therefore, the total contribution of each digit \(d\) to the sum of all 3-digit numbers is \(100d + 10d + d = 111d\).

4. **Summing over all digits**:
    - We need to sum this contribution for all digits from \(1\) to \(9\).
    - The sum of the digits from \(1\) to \(9\) is:
      \[
      1 + 2 + 3 + \cdots + 9 = \frac{9 \cdot (9 + 1)}{2} = 45
      \]
    - Therefore, the total sum of all contributions is:
      \[
      111 \cdot (1 + 2 + 3 + \cdots + 9) = 111 \cdot 45
      \]

5. **Final calculation**:
    - Compute the final product:
      \[
      111 \cdot 45 = 111 \cdot (50 - 5) = 111 \cdot 50 - 111 \cdot 5 = 5550 - 555 = 4995
      \]

Conclusion:
\[
\boxed{4995}
\]