49. (感洏㢦1) Let $f(n)$ be a function defined on the set of positive integers, with values also in this set. If for all positive integers $n, m, f(f(n)+f(m))=n+n$ holds, find all possible values of $f(1938)$.

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49. (感洏㢦1) Let $f(n)$ be a function defined on the set of positive integers, with values also in this set. If for all positive integers $n, m, f(f(n)+f(m))=n+n$ holds, find all possible values of $f(1938)$.
19. (Jiangxi 1)

Solution: We prove that for all $n$, $f(n)=n$, so $f(1983)=1988$ is the only possible value.

Let $r=f(1), s=f(2)$. For simplicity, we denote the equations as follows:
$$
\begin{array}{l}
f(f(n)+f(m))=n+m, \quad (1) \\
f(1)=r, \quad (2) \\
f(2)=s. \quad (3)
\end{array}
$$

From (1) and (3), $f(2s)=4$.
From (1) and (2), $f(2r)=2$.
From (1) and (5), $f(4)=4r$.
From (1), (2), and (6), $f(5r)=5$.
From (1), (2), and (3), $f(s+r)=3$. \quad (8)

From (1) and (4), $f(8)=4s$.
From (1), (7), and (8), $f(8)=6r+s$.
Thus, $4s=6r+s$. Hence, $s=2r$. We will prove by induction:
For each $n \geqslant 4$,
$$
\begin{array}{l}
f(nr)=n, \\
f(n)=nr.
\end{array}
$$

First, from (1), (5), and (8), we have $f(5)=3r+s=5r$. Therefore, (11) and (12) hold for $n=4,5$: (see (4), (6), (7)).
Second, assume (11) holds for $n$ and $n-1$. Then
$$
f((n-1)r)=n-1.
$$

Using (1), (5), and (13), we get $f(n+1)=(n-1)r+2r=(n+1)r$. Also,
$$
f(n-1)=(n-1)r.
$$

Thus, from (14), (1), and (3), we have $f((n+1)r)=n+1$. This completes the induction. Since $4r \geqslant 4$, from (12) we have $f(4r)=4r^2$. And from (11), $f(4r)=4$. Therefore, $r=1$. From (2), (3), (8), and (11), we know that for all $n$, $f(n)=n$.