Find all the pairs of real numbers $(x,y)$ that are solutions of the system:
$(x^{2}+y^{2})^{2}-xy(x+y)^{2}=19 $ 
$| x - y | = 1$
We are given the system of equations:
\[
(x^2 + y^2)^2 - xy(x + y)^2 = 19
\]
\[
|x - y| = 1
\]

We will consider two cases based on the absolute value condition.

### Case 1: \( x - y = 1 \)
1. Substitute \( x = y + 1 \) into the first equation:
   \[
   ((y + 1)^2 + y^2)^2 - (y + 1)y((y + 1) + y)^2 = 19
   \]
2. Simplify the expression:
   \[
   ((y^2 + 2y + 1) + y^2)^2 - (y^2 + y)((2y + 1)^2) = 19
   \]
   \[
   (2y^2 + 2y + 1)^2 - (y^2 + y)(4y^2 + 4y + 1) = 19
   \]
3. Expand and simplify:
   \[
   (2y^2 + 2y + 1)^2 = 4y^4 + 8y^3 + 10y^2 + 4y + 1
   \]
   \[
   (y^2 + y)(4y^2 + 4y + 1) = 4y^4 + 8y^3 + 5y^2 + y
   \]
   \[
   4y^4 + 8y^3 + 10y^2 + 4y + 1 - (4y^4 + 8y^3 + 5y^2 + y) = 19
   \]
   \[
   5y^2 + 3y + 1 = 19
   \]
4. Solve the quadratic equation:
   \[
   5y^2 + 3y - 18 = 0
   \]
   Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{-3 \pm \sqrt{9 + 360}}{10} = \frac{-3 \pm 21}{10}
   \]
   \[
   y = \frac{18}{10} = 1.8 \quad \text{or} \quad y = \frac{-24}{10} = -2.4
   \]
5. Find corresponding \( x \):
   \[
   x = y + 1 \implies x = 2.8 \quad \text{or} \quad x = -1.4
   \]

### Case 2: \( y - x = 1 \)
1. Substitute \( y = x + 1 \) into the first equation:
   \[
   (x^2 + (x + 1)^2)^2 - x(x + 1)(x + (x + 1))^2 = 19
   \]
2. Simplify the expression:
   \[
   (x^2 + x^2 + 2x + 1)^2 - x(x + 1)(2x + 1)^2 = 19
   \]
   \[
   (2x^2 + 2x + 1)^2 - x(x + 1)(4x^2 + 4x + 1) = 19
   \]
3. Expand and simplify:
   \[
   (2x^2 + 2x + 1)^2 = 4x^4 + 8x^3 + 10x^2 + 4x + 1
   \]
   \[
   x(x + 1)(4x^2 + 4x + 1) = 4x^4 + 8x^3 + 5x^2 + x
   \]
   \[
   4x^4 + 8x^3 + 10x^2 + 4x + 1 - (4x^4 + 8x^3 + 5x^2 + x) = 19
   \]
   \[
   5x^2 + 3x + 1 = 19
   \]
4. Solve the quadratic equation:
   \[
   5x^2 + 3x - 18 = 0
   \]
   Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{-3 \pm \sqrt{9 + 360}}{10} = \frac{-3 \pm 21}{10}
   \]
   \[
   x = \frac{18}{10} = 1.8 \quad \text{or} \quad x = \frac{-24}{10} = -2.4
   \]
5. Find corresponding \( y \):
   \[
   y = x + 1 \implies y = 2.8 \quad \text{or} \quad y = -1.4
   \]

### Conclusion
The solutions to the system are:
\[
(x, y) = (2.8, 1.8), (-1.4, -2.4), (1.8, 2.8), (-2.4, -1.4)
\]

The final answer is \(\boxed{(2.8, 1.8), (-1.4, -2.4), (1.8, 2.8), (-2.4, -1.4)}\)