6-124 Let $f(n)$ be a function defined on all positive integers and taking positive integer values. For all positive integers $m, n$, we have
$$f(f(m)+f(n))=m+n .$$

Find all possible values of $f(1988)$.
[Solution] Let $f(1)=t$. If $f(m)=n$, then
$$f(2 n)=f(f(m)+f(m))=2 m$$

In particular, when $m=1, n=t$,
$$f(2 t)=2$$

We prove that $t=1$. Otherwise, let
$$t=b+1, b \in \mathbb{N}$$

Suppose $f(b)=c, c \in \mathbb{N}$, then $f(2 c)=2 b$, and
$$2 c+2 t=f(f(2 c)+f(2 t))=f(2 b+2)=f(2 t)=2$$

Thus, $t+c=1$, which is a contradiction. Therefore, $f(1)=1$. If
$$f(n)=n$$

then $f(n+1)=f(f(n)+f(1))=n+1$. So (4) holds for all natural numbers $n$, in particular,
$$f(1988)=1988$$