Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows: For every $x, y \in \mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$. This condition may in turn be rewritten in the following form: If $g(x) \neq g(y)$, then the number $x+y$ lies (non-strictly) between $g(x)$ and $g(y)$. Notice here that the function $g_{1}(x)=-g(-x)$ also satisfies $(*)$, since
$$\begin{aligned} g_{1}(x) \neq g_{1}(y) \Longrightarrow & g(-x) \neq g(-y) \Longrightarrow \quad-(x+y) \text { lies between } g(-x) \text { and } g(-y) \\ & \Longrightarrow \quad x+y \text { lies between } g_{1}(x) \text { and } g_{1}(y) \end{aligned}$$
On the other hand, the relation we need to prove reads now as
$$g(x) \leqslant g(y) \quad \text { whenever } x < y.$$
Similarly, if $X > 2x$, then $g$ attains at most two values on $[x ; X-x)$ - namely, $X$ and, possibly, some $Y < X$ and hence by (*) we get $X \leqslant a + x \leqslant g(a)$. Now, for any $b \in (X-x ; x)$ with $g(b) \neq X$ we similarly get $b + x \leqslant g(b)$. Therefore, the number $a + b$ (which is smaller than each of $a + x$ and $b + x$) cannot lie between $g(a)$ and $g(b)$, which by (*) implies that $g(a) = g(b)$. Hence $g$ may attain only two values on $(X-x ; x]$, namely $X$ and $g(a) > X$. To prove the second claim, notice that $g_{1}(-x) = -X - X$. Passing back to $g$, we get what we need. Lemma 2. If $X > 2x$, then $g$ is constant on $(x ; X-x)$. Proof. Again, it suffices to prove the first claim only. Assume, for the sake of contradiction, that there exist $a, b \in (X-x ; x)$ with $g(a) \neq g(b)$; by Lemma 1, we may assume that $g(a) = X$ and $Y = g(b) > X$. Notice that $\min \{X - a, X - b\} > X - x$, so there exists a $u \in (X - x ; x)$ such that $u > 2x$. Applying Lemma 2 to $a$ in place of $x$, we obtain that $g$ is constant on $(a, Y - a)$. By (2) again, we have $x \leqslant Y - b < g(y)$ for some $x < y$. Denote $X = g(x)$ and $Y = g(y)$; by (*), we have $X \geqslant x + y \geqslant Y$, so $Y - y \leqslant x < y \leqslant X - x$, and hence $(Y - y ; y) \cap (x ; X - x) = (x, y) \neq \varnothing$. On the other hand, since $Y - y < y$ and $x < X - x$, Lemma 3 shows that $g$ should attain a constant value $X$ on $(x ; X - x)$ and a constant value $Y \neq X$ on $(Y - y ; y)$. Since these intervals overlap, we get the final contradiction.