Determine all functions $f:\mathbb{R}\to\mathbb{R}$ for which there exists a function $g:\mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)=\lfloor g(x+y)\rfloor$ for all real numbers $x$ and $y$.

[i]Emil Vasile[/i]
1. **Initial Setup and Functional Equation:**
   We are given that there exists a function \( g: \mathbb{R} \to \mathbb{R} \) such that for all real numbers \( x \) and \( y \),
   \[
   f(x) + f(y) = \lfloor g(x+y) \rfloor.
   \]
   Let's denote this equation as \( P(x, y) \).

2. **Special Case Analysis:**
   Consider \( P(x, 0) \):
   \[
   f(x) + f(0) = \lfloor g(x) \rfloor.
   \]
   Similarly, consider \( P(0, y) \):
   \[
   f(0) + f(y) = \lfloor g(y) \rfloor.
   \]
   By comparing these two equations, we see that:
   \[
   f(x) + f(0) = \lfloor g(x) \rfloor \quad \text{and} \quad f(0) + f(y) = \lfloor g(y) \rfloor.
   \]

3. **Additivity of \( f \):**
   Now consider \( P(x, y) \) again:
   \[
   f(x) + f(y) = \lfloor g(x+y) \rfloor.
   \]
   By comparing \( P(x+y, 0) \) and \( P(x, y) \), we get:
   \[
   f(x+y) + f(0) = \lfloor g(x+y) \rfloor.
   \]
   Therefore, we have:
   \[
   f(x+y) + f(0) = f(x) + f(y).
   \]
   Let \( h(x) = f(x) - f(0) \). Then:
   \[
   h(x+y) = h(x) + h(y).
   \]
   This shows that \( h \) is an additive function.

4. **Properties of \( h \):**
   From \( P(0, 0) \), we have:
   \[
   2f(0) = \lfloor g(0) \rfloor.
   \]
   This implies that \( 2f(0) \) is an integer, so \( f(0) \) must be a rational number of the form \( \frac{n}{2} \) for some integer \( n \).

5. **Integer Constraints:**
   From \( P(x, x) \), we get:
   \[
   2f(x) = \lfloor g(2x) \rfloor.
   \]
   Since \( 2f(x) \) must be an integer, it follows that \( f(x) \) must be of the form \( \frac{k}{2} \) for some integer \( k \).

6. **Linearity of \( h \):**
   Since \( h \) is additive and \( 2h(x) \) is an integer for all \( x \), \( h \) must be a linear function. Let \( h(x) = ax \). Then:
   \[
   2ax \in \mathbb{Z} \quad \text{for all} \quad x.
   \]
   This implies that \( a = 0 \) because if \( a \neq 0 \), there would exist some \( x \) such that \( 2ax \) is not an integer, leading to a contradiction.

7. **Conclusion:**
   Since \( h(x) = 0 \), we have \( f(x) = f(0) \) for all \( x \). Therefore, \( f \) is a constant function. Let \( f(x) = c \). From the integer constraint, \( c = \frac{n}{2} \) for some integer \( n \).

   Thus, the function \( f \) must be of the form:
   \[
   f(x) = \frac{n}{2} \quad \text{for some} \quad n \in \mathbb{Z}.
   \]

The final answer is \( \boxed{ f(x) = \frac{n}{2} } \) for some \( n \in \mathbb{Z} \).