26 Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have
$$
f(f(x)+y)=2 x+f(f(y)-x) .
$$
For each real number $c$, the function $f(x)=x+c$ is a solution to the functional equation.
Next, we prove that there are no other solutions.
The method is to prove that the equation is of the form $f(X)=X+c$, where $X$ is a function that takes all real numbers.
First, we prove that $f$ is surjective. In fact, let $y=-f(x)$, then
$$
f(0)=2 x+f(f(-f(x))-x) .
$$

That is,
$$
f(f(-f(x))-x)=-2 x+f(0),
$$

Since $-2 x+f(0)$ is a surjective function on $\mathbf{R}$, $f$ is a surjective function on $\mathbf{R}$.
Next, we prove the conclusion we want. From the above, we know that there exists $a \in \mathbf{R}$ such that $f(a)=0$. Let $x=a$ in the equation, then
$$
f(y)=2 a+f(f(y)-a) \Rightarrow f(y)-a=f(f(y)-a)+a .
$$

Since $f$ is surjective, for any $x \in \mathbf{R}$, there exists $y \in \mathbf{R}$ such that $x=f(y)-a$, thus $x=f(x)+a$, hence $f(x)=x-a$.