Find all the pairs $(a,b)$ of integer numbers such that:
$\triangleright$ $a-b-1|a^2+b^2$
$\triangleright$   $\frac{a^2+b^2}{2ab-1}=\frac{20}{19}$
1. Given the conditions:
   \[
   a - b - 1 \mid a^2 + b^2
   \]
   and
   \[
   \frac{a^2 + b^2}{2ab - 1} = \frac{20}{19}
   \]

2. From the second condition, we can rewrite it as:
   \[
   19(a^2 + b^2) = 20(2ab - 1)
   \]
   Simplifying, we get:
   \[
   19a^2 + 19b^2 = 40ab - 20
   \]
   \[
   19a^2 + 19b^2 - 40ab = -20
   \]
   \[
   19a^2 + 19b^2 - 40ab + 20 = 0
   \]

3. Let \( a - b = c \). Then \( a = b + c \). Substituting \( a = b + c \) into the equation:
   \[
   19(b + c)^2 + 19b^2 - 40(b + c)b + 20 = 0
   \]
   Expanding and simplifying:
   \[
   19(b^2 + 2bc + c^2) + 19b^2 - 40b^2 - 40bc + 20 = 0
   \]
   \[
   19b^2 + 38bc + 19c^2 + 19b^2 - 40b^2 - 40bc + 20 = 0
   \]
   \[
   38b^2 - 2bc + 19c^2 + 20 = 0
   \]

4. Rearranging the terms:
   \[
   19c^2 + 20 = 2bc + 2b^2
   \]

5. From the first condition, \( c - 1 \mid c^2 + 19c^2 + 20 \):
   \[
   c - 1 \mid 20c^2 + 20
   \]
   Since \( c - 1 \mid c^2 - 1 \), we know:
   \[
   c - 1 \mid 20
   \]
   Therefore, \( c - 1 \) can be one of the divisors of 20: \( \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 \).

6. Testing the possible values of \( c \):
   - For \( c = 2 \):
     \[
     a = b + 2
     \]
     Substituting into the equation:
     \[
     \frac{(b + 2)^2 + b^2}{2(b + 2)b - 1} = \frac{20}{19}
     \]
     Simplifying:
     \[
     \frac{b^2 + 4b + 4 + b^2}{2b^2 + 4b - 1} = \frac{20}{19}
     \]
     \[
     \frac{2b^2 + 4b + 4}{2b^2 + 4b - 1} = \frac{20}{19}
     \]
     Cross-multiplying:
     \[
     19(2b^2 + 4b + 4) = 20(2b^2 + 4b - 1)
     \]
     \[
     38b^2 + 76b + 76 = 40b^2 + 80b - 20
     \]
     \[
     2b^2 + 4b - 96 = 0
     \]
     \[
     b^2 + 2b - 48 = 0
     \]
     Solving the quadratic equation:
     \[
     b = \frac{-2 \pm \sqrt{4 + 192}}{2}
     \]
     \[
     b = \frac{-2 \pm 14}{2}
     \]
     \[
     b = 6 \quad \text{or} \quad b = -8
     \]
     Therefore, \( (a, b) = (8, 6) \) or \( (-6, -8) \).

   - For \( c = 6 \):
     \[
     a = b + 6
     \]
     Substituting into the equation:
     \[
     \frac{(b + 6)^2 + b^2}{2(b + 6)b - 1} = \frac{20}{19}
     \]
     Simplifying:
     \[
     \frac{b^2 + 12b + 36 + b^2}{2b^2 + 12b - 1} = \frac{20}{19}
     \]
     \[
     \frac{2b^2 + 12b + 36}{2b^2 + 12b - 1} = \frac{20}{19}
     \]
     Cross-multiplying:
     \[
     19(2b^2 + 12b + 36) = 20(2b^2 + 12b - 1)
     \]
     \[
     38b^2 + 228b + 684 = 40b^2 + 240b - 20
     \]
     \[
     2b^2 + 12b - 704 = 0
     \]
     \[
     b^2 + 6b - 352 = 0
     \]
     Solving the quadratic equation:
     \[
     b = \frac{-6 \pm \sqrt{36 + 1408}}{2}
     \]
     \[
     b = \frac{-6 \pm 38}{2}
     \]
     \[
     b = 16 \quad \text{or} \quad b = -22
     \]
     Therefore, \( (a, b) = (22, 16) \) or \( (-16, -22) \).

7. The final solutions are:
   \[
   (a, b) = (22, 16), (-16, -22), (8, 6), (-6, -8)
   \]

The final answer is \( \boxed{ (22, 16), (-16, -22), (8, 6), (-6, -8) } \).