Example 7 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 $f(1988)$. (1988 Mexican Olympiad Problem)
Let $f(1)=t$. If $f(m)=n$, then $f(2 n)=f[f(m)+f(m)]=2 m$.
In particular, when $m=1, n=t$, we have $f(2 t)=2$.
We prove that $t=1$. Otherwise, let $t=b+1, b \in \mathbf{N}$.
Suppose $f(b)=c, c \in \mathbf{N}$, 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 $t+c=1$. This is a contradiction, so $f(1)=1$.
If $f(n)=n$, then $f(n+1)=f[f(n)+f(1)]=n+1$.
Therefore, for all natural numbers $n$, we have $f(n)=n$.
In particular, $f(1988)=1988$.