2・62 Now there are 1990 piles of stones, each consisting of $1,2, \cdots, 1990$ stones, and in each round, you are allowed to pick any number of piles and throw away the same number of stones from these piles. Question: What is the minimum number of rounds needed to throw away all the stones?
[Solution] After each round, we consider piles with the same number of stones as one group, and empty piles as another group. Suppose at a certain moment there are $n$ groups, and in the next round, the same number of stones is removed from some piles, which originally belonged to $k$ different groups. After the removal, they still belong to $k$ different groups, because the number of stones remaining in different groups after removal is still different. The rest of the piles then belong to at least $n-k$ different groups. Therefore, the total number of groups will not be less than $\max \{k, n-k\}$.

This means that after each round, the number of groups can decrease by at most half of the original number. Thus, the rate at which the number of stone groups decreases cannot be faster than the following standard sequence:
$995,498,249,125,63,32,16,8,4,2,1$, which means it takes 11 rounds.
On the other hand, there indeed exists a method to clear all the stones in 11 rounds: In the first round, remove 995 stones from each pile that has at least 995 stones, leaving the number of stones in each pile as follows:
$$
995,994, \cdots, 3,2,1,0 \text {. }
$$

In the second round, remove 498 stones from each pile that has at least 498 stones, leaving the number of stones in each pile as follows:
$$
497,496, \cdots, 3,2,1,0 \text {. }
$$

Continuing in this manner, let $n$ take each value in the standard sequence in turn, and each time remove $\left[\frac{n+1}{n}\right]$ stones from each pile that has at least $\frac{n}{2}$ stones, so that the number of stones remaining in each pile is only as follows:
$$
n-1, n-2, \cdots, 1,0 \text {. }
$$

Thus, it takes only 11 rounds to clear all the stones.