Find a pair of relatively prime four digit natural numbers $A$ and $B$ such that for all natural numbers $m$ and $n$, $\vert A^m -B^n \vert \ge 400$.
To find a pair of relatively prime four-digit natural numbers \( A \) and \( B \) such that for all natural numbers \( m \) and \( n \), \( |A^m - B^n| \ge 400 \), we can proceed as follows:

1. **Choose \( A \) and \( B \) such that they are relatively prime:**
   - Let \( A = 3125 = 5^5 \).
   - Let \( B = 9376 \).

2. **Verify that \( A \) and \( B \) are relatively prime:**
   - The prime factorization of \( A \) is \( 5^5 \).
   - The prime factorization of \( B \) does not include the prime number 5.
   - Therefore, \( \gcd(3125, 9376) = 1 \), confirming that \( A \) and \( B \) are relatively prime.

3. **Check the condition \( |A^m - B^n| \ge 400 \) for all natural numbers \( m \) and \( n \):**
   - Consider the congruences modulo 10000:
     \[
     A \equiv 3125 \pmod{10000}
     \]
     \[
     A^2 \equiv 3125^2 \equiv 9765625 \equiv 5625 \pmod{10000}
     \]
     \[
     A^3 \equiv 3125^3 \equiv 17578125 \equiv 625 \pmod{10000}
     \]
     \[
     A^4 \equiv 3125^4 \equiv 1953125 \equiv 625 \pmod{10000}
     \]
   - For \( B \):
     \[
     B \equiv 9376 \pmod{10000}
     \]
     \[
     B^2 \equiv 9376^2 \equiv 87909376 \equiv 9376 \pmod{10000}
     \]
   - Notice that \( B^n \equiv 9376 \pmod{10000} \) for all \( n \ge 1 \).

4. **Evaluate \( |A^m - B^n| \) modulo 10000:**
   - For \( m = 1 \):
     \[
     |A^1 - B^n| = |3125 - 9376| = 6251
     \]
   - For \( m = 2 \):
     \[
     |A^2 - B^n| = |5625 - 9376| = 3751
     \]
   - For \( m \ge 3 \):
     \[
     |A^m - B^n| = |625 - 9376| = 8751
     \]

5. **Verify the condition \( |A^m - B^n| \ge 400 \):**
   - For all \( m \ge 1 \) and \( n \ge 1 \), the minimum value of \( |A^m - B^n| \) is 3751, which is greater than 400.

Thus, the pair \( (A, B) = (3125, 9376) \) satisfies the given condition.

The final answer is \( \boxed{ (A, B) = (3125, 9376) } \).