If $a,b,x,y$ are real numbers then find all solutions of $a,b$ such that $\left \lfloor ax+by \right \rfloor +  \left \lfloor bx+ay \right \rfloor = (a+b) \left \lfloor x+y \right \rfloor$
To solve the given problem, we need to find all real numbers \(a\) and \(b\) such that the equation 
\[
\left \lfloor ax+by \right \rfloor + \left \lfloor bx+ay \right \rfloor = (a+b) \left \lfloor x+y \right \rfloor
\]
holds for all real numbers \(x\) and \(y\).

1. **Substitute specific values for \(x\) and \(y\):**
   - Let \(x = y = 0\):
     \[
     \left \lfloor a \cdot 0 + b \cdot 0 \right \rfloor + \left \lfloor b \cdot 0 + a \cdot 0 \right \rfloor = (a+b) \left \lfloor 0 + 0 \right \rfloor
     \]
     \[
     \left \lfloor 0 \right \rfloor + \left \lfloor 0 \right \rfloor = (a+b) \left \lfloor 0 \right \rfloor
     \]
     \[
     0 + 0 = (a+b) \cdot 0
     \]
     \[
     0 = 0
     \]
     This equation is always true and does not provide any new information about \(a\) and \(b\).

2. **Consider \(x = 1\) and \(y = -1\):**
   \[
   \left \lfloor a \cdot 1 + b \cdot (-1) \right \rfloor + \left \lfloor b \cdot 1 + a \cdot (-1) \right \rfloor = (a+b) \left \lfloor 1 + (-1) \right \rfloor
   \]
   \[
   \left \lfloor a - b \right \rfloor + \left \lfloor b - a \right \rfloor = (a+b) \left \lfloor 0 \right \rfloor
   \]
   \[
   \left \lfloor a - b \right \rfloor + \left \lfloor b - a \right \rfloor = 0
   \]
   Since \(\left \lfloor a - b \right \rfloor\) and \(\left \lfloor b - a \right \rfloor\) are integers, the only way their sum can be zero is if both are zero:
   \[
   \left \lfloor a - b \right \rfloor = 0 \quad \text{and} \quad \left \lfloor b - a \right \rfloor = 0
   \]
   This implies:
   \[
   0 \leq a - b < 1 \quad \text{and} \quad 0 \leq b - a < 1
   \]
   The only solution to these inequalities is \(a = b\).

3. **Substitute \(a = b\) into the original equation:**
   \[
   \left \lfloor ax + by \right \rfloor + \left \lfloor bx + ay \right \rfloor = (a+b) \left \lfloor x+y \right \rfloor
   \]
   \[
   \left \lfloor ax + ay \right \rfloor + \left \lfloor ax + ay \right \rfloor = 2a \left \lfloor x+y \right \rfloor
   \]
   \[
   2 \left \lfloor a(x + y) \right \rfloor = 2a \left \lfloor x+y \right \rfloor
   \]
   This equation holds if and only if \(a\) is an integer. Therefore, \(a\) and \(b\) must be integers.

4. **Check the possible integer values for \(a\) and \(b\):**
   - If \(a = b = 0\):
     \[
     \left \lfloor 0 \cdot x + 0 \cdot y \right \rfloor + \left \lfloor 0 \cdot x + 0 \cdot y \right \rfloor = (0+0) \left \lfloor x+y \right \rfloor
     \]
     \[
     0 + 0 = 0
     \]
     This is true for all \(x\) and \(y\).

   - If \(a = b = 1\):
     \[
     \left \lfloor 1 \cdot x + 1 \cdot y \right \rfloor + \left \lfloor 1 \cdot x + 1 \cdot y \right \rfloor = (1+1) \left \lfloor x+y \right \rfloor
     \]
     \[
     2 \left \lfloor x + y \right \rfloor = 2 \left \lfloor x+y \right \rfloor
     \]
     This is also true for all \(x\) and \(y\).

Therefore, the solutions are \(a = b = 0\) or \(a = b = 1\).

The final answer is \( \boxed{ a = b = 0 } \) or \(a = b = 1\).