China National Olympiad

Let $a = 2001$. Consider the set $A$ of all pairs of integers $(m, n)$ with $n \neq 0$ such that:
1. $m < 2a$,
2. $2n \mid (2am - m^2 + n^2)$,
3. $n^2 - m^2 + 2mn \leq 2a(n - m)$.

For $(m, n) \in A$, let
$$f(m, n) = \frac{2am - m^2 - mn}{n}.$$
We need to determine the maximum and minimum values of $f$.

### Minimum Value of $f$
From condition (ii), we have:
$$\frac{2am - m^2 + n^2}{2n} = \ell \in \mathbb{Z} \implies n(2\ell - n) = m(2a - m).$$
Thus, $m \equiv n \pmod{2}$.

Using condition (iii):
$$\ell = \frac{(2am - 2mn) + (n^2 - m^2 + 2mn)}{2n} \leq \frac{(2am - 2mn) + 2a(n - m)}{2n} = a - m.$$

From this, we have:
$$2\ell - n < 2\ell \leq 2a - 2m < 2a - m.$$
Using this and the previous equation, we conclude $n > m$. Also, from condition (iii):
$$2mn \leq 2a(n - m) - (n^2 - m^2) \implies (n - m)(2a - n - m).$$

Thus, $2a - n - m > 0$. Therefore, $f(m, n) = \frac{m(2a - m - n)}{n} > 0$.

Hence, $f(m, n) = 2\ell - (m + n) \equiv 0 \pmod{2}$ and $f(m, n) > 0$. Thus, we conclude $f(m, n) \geq 2$ and the equality holds for $f(2, 2000) = 2$.

### Maximum Value of $f$
Consider:
$$f(n - 2, n) = \frac{2a(n - 2) - (n - 2)^2 - n(n - 2)}{n} = 2a + 6 - 2\left(n + \frac{4004}{n}\right).$$

To maximize this, we need to minimize $n + \frac{4004}{n}$. Choosing $n \mid 4004$ such that $n$ and $\frac{4004}{n}$ are as close as possible, we find $n = 52$ and $m = 50$ satisfy the conditions. Therefore, $f(50, 52) = 3750$.

Since $n > m$ and $m \equiv n \pmod{2}$, it suffices to prove that for $n \geq m + 4$, $f(m, n) < 3750$:
$$f(m, n) = \frac{(2a - m)m}{n} - m \leq \frac{(2a - m)m}{m + 4} - m = 3998 - \left(2(m + 4) + \frac{16024}{m + 4}\right) \leq 3998 - 2\sqrt{32048} < 3640.$$

Thus, the maximum value of $f$ is $3750$.

### Conclusion
The minimum value of $f(m, n)$ is $2$, and the maximum value is $3750$.

The answer is: $\boxed{2 \text{ and } 3750}$.