Problem 1654

Proof:
The idea of the proof is to find a constant such that for any $n \in \mathbf{N}^{*}$, the product of $\frac{A(n)}{B(n)}$ and this constant is a positive integer, thereby deducing that $f$ is a bounded function.

First, we prove a lemma: For any non-negative real numbers $x, y$,
$$[2 x]+[2 y] \geqslant[x]+[x+y]$$

In fact, let $x=m+\alpha, y=n+\beta$, where $m, n \in \mathbf{N}$, and $\alpha, \beta \in[0,1)$, then (2) is equivalent to proving

that is,
$$\begin{array}{c} 2 m+[2 \alpha]+2 n+[2 \beta] \geqslant m+(m+n)+[\alpha+\beta] \\ n+[2 \alpha]+[2 \beta] \geqslant[\alpha+\beta] . \end{array}$$

Since $[\alpha+\beta] \leqslant 1$, and when $[\alpha+\beta]=1$, at least one of $\alpha, \beta$ is not less than $\frac{1}{2}$, thus in this case $[2 \alpha]+[2 \beta] \geqslant 1$. Therefore, (3) holds. The lemma is proved.

Returning to the original problem, we first prove: For any $n \in \mathbf{N}^{*}$, the number $\frac{(2 n)!\cdot 2000!}{n!(n+1000)!}$ is a positive integer. To do this, we only need to prove that for any prime $p$, the power of $p$ in the prime factorization of $(2 n)!\cdot 2000!$ is not less than the power of $p$ in the prime factorization of $n!(n+1000)!$. Using property 5, we only need to prove:
$$\sum_{k=1}^{+\infty}\left(\left[\frac{2 n}{p^{k}}\right]+\left[\frac{2000}{p^{k}}\right]\right)-\left(\left[\frac{n}{p^{k}}\right]+\left[\frac{n+1000}{p^{k}}\right]\right) \geqslant 0$$

If we let $x=\frac{n}{p^{k}}, y=\frac{1000}{p^{k}}$, using (2) we know that inequality (4) holds, so $\frac{(2 n)!\cdot 2000!}{n!(n+1000)!} \in \mathbf{N}^{*}$, i.e.,
$$\frac{A(n)}{B(n)} \cdot 2000!\in \mathbf{N}^{*}$$

The above discussion shows that for any $n \in \mathbf{N}^{*}$, we have $f(n) \mid 2000$!. Noting that $f(n)$ is 1 or a prime number, and by direct verification, 1999 is a prime number, so for any $n \in \mathbf{N}^{*}$, we have $f(n) \leqslant 1999$. Therefore, $f(n)$ is a bounded function.

Furthermore, when $n=999$, we know that
$$B(n)=1999 \cdot(999!)$$

At this time, $f(n)=1999$. Hence, the maximum value of $f(n)$ is 1999.