Question 227: There are 2021 piles of stones, with the number of stones in each pile being 1, 2, ..., 2021 respectively. The following operation can be performed: select any number of piles and remove the same number of stones from each selected pile. How many operations are required at minimum to remove all the stones?
Question 227, Solution: Suppose there are $\mathrm{n}$ piles of stones, with the number of stones in each pile being distinct (according to the problem, piles with the same number of stones can be treated as one pile at the same time). If, during an operation, $\mathrm{k}$ piles are taken, after the operation, these $\mathrm{k}$ piles can decrease by at most one pile (i.e., one pile of stones is completely taken away), and the remaining $n-k$ piles remain unchanged. Even if some of the $\mathrm{k}$ piles taken become the same as the $n-k$ piles, the number of piles is still $\geq \max (n-k, k-1) \geq\left\lceil\frac{(n-k)+(k-1)}{2}\right\rceil=\left[\frac{n}{2}\right]$.
Let $\mathrm{f}(\mathrm{n})$ be the minimum number of operations required for $\mathrm{n}$ piles of stones. According to the above analysis, we have:
$$
f(n) \geq f\left(\left[\frac{n}{2}\right]\right)+1
$$

Also, note that $\mathrm{f}(1)=1$. By mathematical induction, it is easy to see that $\mathrm{f}(\mathrm{n}) \geq\left[\log _{2} \mathrm{n}\right]+1$.
On the other hand, if the number of stones in the $\mathrm{n}$ piles are $1, 2, \cdots, n$, then these $\mathrm{n}$ numbers, when represented in binary, have at most $\left[\log _{2} \mathrm{n}\right]+1$ digits. By performing $\left[\log _{2} \mathrm{n}\right]+1$ operations, all stones can be taken away: in the $\mathrm{i}(1 \leq \mathrm{i} \leq \left.\left[\log _{2} n\right]+1\right)$-th operation, take away $2^{\left[\log _{2} n\right]+i-1}$ stones from all piles with at least $2^{\left[\log _{2} n\right]+i-1}$ stones.
In particular, when $n=2021$, at least 11 operations are required.