6-12 $f(n)$ is a function defined on the set of positive integers, taking non-negative integer values, and for all $m, n$ we have:
$$
\begin{array}{l}
f(m+n)-f(m)-f(n)=0 \text { or } 1 ; \\
f(2)=0, f(3)>0, f(9999)=3333 .
\end{array}
$$

Try to find: $f(1982)$.
[Solution] Let $m=n=1$, then $f(2)=2 f(1)+(0$ or 1$)$. Since $f(2)=0$, we have $2 f(1)+(0$ or 1$)=0$. Also, $f(1)$ is an integer, so $f(1)=0$.
Let $m=2, n=1$, then
$f(3)=f(2)+f(1)+(0$ or 1$)$,
$f(3)=(0$ or 1$)$, and since $f(3)>0$, we have $f(3)=1$.
Let $m=n+1$, then
\[
\begin{aligned}
f(3 n+3) & =f(3 n)+f(3)+(0 \text{ or } 1) \\
& =f(3 n)+(1 \text{ or } 2),
\end{aligned}
\]

Therefore, $f(3(n+1)) \geqslant f(3 n)+1$.
By induction, we get $f(3 n) \geqslant n$.
However, if $f(3 n)>n$ for some $n$, then it holds for all larger numbers.
Since $f(9999)=3333$, for $n \leqslant 3333$,

we have $f(3 n)=n$ (where $n \leqslant 3333$).
\[
\begin{array}{l}
\text{Finally, } 1982=f(3 \times 1982) \geqslant f(2 \times 1982)+f(1982) \\
\geqslant 3 f(1982) \text{.} \\
\end{array}
\]

Thus, $f(1982) \leqslant 1982 / 3<661$.
Also, $f(1982) \geqslant f(1980)+f(2)=660$,
so $f(1982)=660$.