A is a two-digit number and B is a three-digit number such that A increased by B% equals B reduced by A%. Find all possible pairs (A, B).
1. We start with the given condition: \( A \) increased by \( B\% \) equals \( B \) reduced by \( A\% \). This can be written as:
   \[
   A \left(1 + \frac{B}{100}\right) = B \left(1 - \frac{A}{100}\right)
   \]

2. Simplify the equation:
   \[
   A \left(1 + \frac{B}{100}\right) = B \left(1 - \frac{A}{100}\right)
   \]
   \[
   A + \frac{AB}{100} = B - \frac{AB}{100}
   \]

3. Combine like terms:
   \[
   A + \frac{AB}{100} + \frac{AB}{100} = B
   \]
   \[
   A + \frac{2AB}{100} = B
   \]
   \[
   A + \frac{AB}{50} = B
   \]

4. Rearrange the equation to isolate terms involving \( A \) and \( B \):
   \[
   B - A = \frac{AB}{50}
   \]
   \[
   50(B - A) = AB
   \]
   \[
   AB + 50A - 50B = 0
   \]

5. Factor the equation:
   \[
   (50 - A)(B + 50) = 2500
   \]

6. Since \( 50 - A \) must be a two-digit number (and \( A \) can't be zero), the possible values for \( 50 - A \) are:
   \[
   50 - A \in \{10, 20, 25\}
   \]

7. Check each possibility:
   - For \( 50 - A = 10 \):
     \[
     A = 40
     \]
     \[
     (50 - 40)(B + 50) = 2500
     \]
     \[
     10(B + 50) = 2500
     \]
     \[
     B + 50 = 250
     \]
     \[
     B = 200
     \]
     This gives the pair \( (A, B) = (40, 200) \).

   - For \( 50 - A = 20 \):
     \[
     A = 30
     \]
     \[
     (50 - 30)(B + 50) = 2500
     \]
     \[
     20(B + 50) = 2500
     \]
     \[
     B + 50 = 125
     \]
     \[
     B = 75
     \]
     This does not satisfy the conditions as \( B \) must be a three-digit number.

   - For \( 50 - A = 25 \):
     \[
     A = 25
     \]
     \[
     (50 - 25)(B + 50) = 2500
     \]
     \[
     25(B + 50) = 2500
     \]
     \[
     B + 50 = 100
     \]
     \[
     B = 50
     \]
     This does not satisfy the conditions as \( B \) must be a three-digit number.

8. Therefore, the only valid solution is:
   \[
   (A, B) = (40, 200)
   \]

The final answer is \(\boxed{(40, 200)}\).