Find all natural numbers with the property that, when the first digit is moved to the end, the resulting number is $\dfrac{7}{2}$ times the original one.
1. Let \( n \) be the original number. We can express \( n \) in the form \( n = A + 10^k b \), where \( b \) is the first digit, \( k+1 \) is the number of digits in \( n \), and \( A \) is the remaining part of the number after removing the first digit.

2. When the first digit \( b \) is moved to the end, the resulting number is \( 10A + b \).

3. According to the problem, this new number is \( \frac{7}{2} \) times the original number. Therefore, we have the equation:
   \[
   10A + b = \frac{7}{2} (A + 10^k b)
   \]

4. To eliminate the fraction, multiply both sides by 2:
   \[
   2(10A + b) = 7(A + 10^k b)
   \]
   \[
   20A + 2b = 7A + 7 \cdot 10^k b
   \]

5. Rearrange the terms to isolate \( A \):
   \[
   20A - 7A = 7 \cdot 10^k b - 2b
   \]
   \[
   13A = b(7 \cdot 10^k - 2)
   \]

6. Since \( b \) is a digit (1 through 9), \( 13 \) must divide \( 7 \cdot 10^k - 2 \). Therefore, we need to find \( k \) such that:
   \[
   7 \cdot 10^k - 2 \equiv 0 \pmod{13}
   \]
   \[
   7 \cdot 10^k \equiv 2 \pmod{13}
   \]

7. We need to find the smallest \( k \) that satisfies this congruence. We can test values of \( k \) to find a solution:
   \[
   10^1 \equiv 10 \pmod{13}
   \]
   \[
   7 \cdot 10 \equiv 70 \equiv 5 \pmod{13}
   \]
   \[
   10^2 \equiv 100 \equiv 9 \pmod{13}
   \]
   \[
   7 \cdot 9 \equiv 63 \equiv 11 \pmod{13}
   \]
   \[
   10^3 \equiv 1000 \equiv 12 \pmod{13}
   \]
   \[
   7 \cdot 12 \equiv 84 \equiv 6 \pmod{13}
   \]
   \[
   10^4 \equiv 10000 \equiv 3 \pmod{13}
   \]
   \[
   7 \cdot 3 \equiv 21 \equiv 8 \pmod{13}
   \]
   \[
   10^5 \equiv 100000 \equiv 4 \pmod{13}
   \]
   \[
   7 \cdot 4 \equiv 28 \equiv 2 \pmod{13}
   \]

   We find that \( k = 5 \) satisfies the congruence.

8. Now, we substitute \( k = 5 \) back into the equation:
   \[
   13A = b(7 \cdot 10^5 - 2)
   \]
   \[
   13A = b(700000 - 2)
   \]
   \[
   13A = b \cdot 699998
   \]

9. Since \( A \) must be an integer, \( b \cdot 699998 \) must be divisible by 13. We find that \( 699998 \equiv 0 \pmod{13} \), so \( b \) can be any digit from 1 to 9.

10. For simplicity, let's choose \( b = 1 \):
    \[
    13A = 1 \cdot 699998
    \]
    \[
    A = \frac{699998}{13} = 53846
    \]

11. Therefore, the original number \( n \) is:
    \[
    n = A + 10^5 b = 53846 + 10^5 \cdot 1 = 153846
    \]

The final answer is \(\boxed{153846}\)