Find all tuples $ (x_1,x_2,x_3,x_4,x_5)$ of positive integers with $ x_1>x_2>x_3>x_4>x_5>0$ and 
$ {\left \lfloor \frac{x_1+x_2}{3} \right \rfloor }^2 + {\left \lfloor \frac{x_2+x_3}{3} \right \rfloor }^2 + {\left \lfloor \frac{x_3+x_4}{3} \right \rfloor }^2 + {\left \lfloor \frac{x_4+x_5}{3} \right \rfloor }^2 = 38.$
1. We start with the given equation:
   \[
   \left\lfloor \frac{x_1 + x_2}{3} \right\rfloor^2 + \left\lfloor \frac{x_2 + x_3}{3} \right\rfloor^2 + \left\lfloor \frac{x_3 + x_4}{3} \right\rfloor^2 + \left\lfloor \frac{x_4 + x_5}{3} \right\rfloor^2 = 38
   \]
   where \(x_1 > x_2 > x_3 > x_4 > x_5 > 0\).

2. We note that the left-hand side (LHS) must be congruent to the right-hand side (RHS) modulo 8:
   \[
   \left\lfloor \frac{x_1 + x_2}{3} \right\rfloor^2 + \left\lfloor \frac{x_2 + x_3}{3} \right\rfloor^2 + \left\lfloor \frac{x_3 + x_4}{3} \right\rfloor^2 + \left\lfloor \frac{x_4 + x_5}{3} \right\rfloor^2 \equiv 38 \pmod{8}
   \]
   Since \(38 \equiv 6 \pmod{8}\), we need the sum of four squares to be congruent to 6 modulo 8.

3. The possible values of squares modulo 8 are 0, 1, and 4. We need to find a combination of these values that sum to 6 modulo 8. The possible combinations are:
   \[
   4 + 1 + 1 + 0 \quad \text{or} \quad 4 + 1 + 1 + 0 \quad \text{(permutations)}
   \]

4. This implies that one of the terms must be 4, and the others must be 1, 1, and 0. Since all terms are non-negative integers, we can deduce:
   \[
   \left\lfloor \frac{x_1 + x_2}{3} \right\rfloor = 2, \quad \left\lfloor \frac{x_2 + x_3}{3} \right\rfloor = 1, \quad \left\lfloor \frac{x_3 + x_4}{3} \right\rfloor = 1, \quad \left\lfloor \frac{x_4 + x_5}{3} \right\rfloor = 0
   \]

5. We now solve for the values of \(x_i\):
   \[
   \left\lfloor \frac{x_1 + x_2}{3} \right\rfloor = 2 \implies 6 \leq x_1 + x_2 < 9
   \]
   \[
   \left\lfloor \frac{x_2 + x_3}{3} \right\rfloor = 1 \implies 3 \leq x_2 + x_3 < 6
   \]
   \[
   \left\lfloor \frac{x_3 + x_4}{3} \right\rfloor = 1 \implies 3 \leq x_3 + x_4 < 6
   \]
   \[
   \left\lfloor \frac{x_4 + x_5}{3} \right\rfloor = 0 \implies 0 \leq x_4 + x_5 < 3
   \]

6. From the inequalities, we can deduce:
   \[
   x_1 + x_2 = 7 \quad \text{(since 6 and 8 are not possible due to integer constraints)}
   \]
   \[
   x_2 + x_3 = 4 \quad \text{(since 3 and 5 are not possible due to integer constraints)}
   \]
   \[
   x_3 + x_4 = 4 \quad \text{(since 3 and 5 are not possible due to integer constraints)}
   \]
   \[
   x_4 + x_5 = 1 \quad \text{(since 0 and 2 are not possible due to integer constraints)}
   \]

7. Solving these equations, we get:
   \[
   x_2 = 4, \quad x_3 = 5, \quad x_4 = 6, \quad x_5 = 1
   \]
   \[
   x_1 = 3 \quad \text{(since } x_1 + x_2 = 7 \text{ and } x_2 = 4)
   \]

8. Therefore, the tuples that satisfy the conditions are:
   \[
   (7, 4, 5, 6, 1)
   \]

The final answer is \( \boxed{ (7, 4, 5, 6, 1) } \).