Setting $x = 0,$ we get
\[f(f(y)) = y\]for all $y.$
In particular,
\[f(f(0)) = 0.\]Setting $y = 0,$ we get
\[f(x + f(0)) = x\]for all $x.$
Since $f(f(0)) = 0,$ this means $f(x) = x + f(0).$
Substituting into the given functional equation, we get
\[x + f(0) + y + f(0) = x + y.\]This gives us $f(0) = 0,$ so $f(x) = x$ for all $x.$
Therefore, there is only $\boxed{1}$ function that works.The answer is: 1
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Find the number of functions $f : \mathbb{R} \to \mathbb{R}$ such that
\[f(x + f(y)) = x + y\]for all real numbers $x$ and $y.$

### Answer:
