, 1989

Denote by $f_n$ the $n$th iterate of $f$, that is, $f_n(x)=\underbrace{f(f(\ldots f}_{n \text{ times }}(x)))$.
Plug $x \rightarrow f_{n+1}(x)$ in (2): since $g\left(f_{n+1}(x)\right)=g\left(f\left(f_n(x)\right)\right)=f_n(x)$,
$$f_{n+2}(x)+f_n(x)=2 f_{n+1}(x),$$
that is,
$$f_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_n(x) .$$
Therefore $f_n(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find
$$f_n(x)-x=n(f(x)-x) .$$
Since $g$ has the same properties as $f$,
$$g_n(x)-x=n(g(x)-x)=-n(f(x)-x) .$$
Finally, $g$ is also increasing, because since $f$ is increasing $g(x)>g(y) \Longrightarrow f(g(x))> f(g(y)) \Longrightarrow x>y$. An induction proves that $f_n$ and $g_n$ are also increasing functions.
Let $x>y$ be real numbers. Since $f_n$ and $g_n$ are increasing,
$$x+n(f(x)-x)>y+n(f(y)-y) \Longleftrightarrow n[(f(x)-x)-(f(y)-y)]>y-x$$
and
$$x-n(f(x)-x)>y-n(f(y)-y) \Longleftrightarrow n[(f(x)-x)-(f(y)-y)]<x-y .$$
Summing it up,
$$|n[(f(x)-x)-(f(y)-y)]|<x-y \quad \text{ for all } n \in \mathbb{Z}_{>0} .$$
Suppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$,
$$|n(a-b)|<x-y,$$
which is false for a sufficiently large $n$. Hence $a=b$, and $f(x)-x$ is a constant $c$ for all $x \in \mathbb{R}$, that is, $f(x)=x+c$.
It is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.