Find all real numbers $a$ such that there exist $f:\mathbb{R} \to \mathbb{R}$ with$$f(x+f(y))=f(x)+a\lfloor y \rfloor $$for all $x,y\in \mathbb{R}$
1. **Given Functional Equation:**
   \[
   f(x + f(y)) = f(x) + a \lfloor y \rfloor
   \]
   for all \( x, y \in \mathbb{R} \).

2. **Substitute \( x = 0 \):**
   \[
   f(f(y)) = f(0) + a \lfloor y \rfloor
   \]
   Let \( c = f(0) \). Then,
   \[
   f(f(y)) = c + a \lfloor y \rfloor
   \]

3. **Substitute \( y = 0 \):**
   \[
   f(x + f(0)) = f(x) + a \lfloor 0 \rfloor
   \]
   Since \( \lfloor 0 \rfloor = 0 \), we have:
   \[
   f(x + c) = f(x)
   \]
   This implies that \( f \) is periodic with period \( c \).

4. **Substitute \( y = 1 \):**
   \[
   f(x + f(1)) = f(x) + a \lfloor 1 \rfloor
   \]
   Since \( \lfloor 1 \rfloor = 1 \), we have:
   \[
   f(x + f(1)) = f(x) + a
   \]

5. **Substitute \( x = -f(1) \):**
   \[
   f(-f(1) + f(1)) = f(-f(1)) + a
   \]
   Simplifying, we get:
   \[
   f(0) = f(-f(1)) + a
   \]
   Since \( f(0) = c \), we have:
   \[
   c = f(-f(1)) + a
   \]
   Therefore,
   \[
   f(-f(1)) = c - a
   \]

6. **Substitute \( y = -1 \):**
   \[
   f(x + f(-1)) = f(x) + a \lfloor -1 \rfloor
   \]
   Since \( \lfloor -1 \rfloor = -1 \), we have:
   \[
   f(x + f(-1)) = f(x) - a
   \]

7. **Substitute \( x = -f(-1) \):**
   \[
   f(-f(-1) + f(-1)) = f(-f(-1)) - a
   \]
   Simplifying, we get:
   \[
   f(0) = f(-f(-1)) - a
   \]
   Since \( f(0) = c \), we have:
   \[
   c = f(-f(-1)) - a
   \]
   Therefore,
   \[
   f(-f(-1)) = c + a
   \]

8. **Consistency Check:**
   From steps 5 and 7, we have:
   \[
   f(-f(1)) = c - a \quad \text{and} \quad f(-f(-1)) = c + a
   \]
   For \( f \) to be consistent, we need \( f(-f(1)) = f(-f(-1)) \). This implies:
   \[
   c - a = c + a
   \]
   Therefore,
   \[
   -a = a \implies a = 0
   \]

Conclusion:
The only value of \( a \) that satisfies the given functional equation is \( a = 0 \).

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