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
$$
f(2 n)=f(f(m)+f(m))=2 m \text{. }
$$

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

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

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

which implies $t+c=1$, 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,
$$
f(1988)=1988 .
$$