7.103 There are 1990 piles of stones, with the number of stones in each pile being $1, 2, \cdots$ 1990. The allowed operation is to 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?
[Solution] After each operation, we group the piles that have the same number of stones. At any time, even if we count the piles with no stones as one group, we always consider there to be $n$ groups. If in the next operation we remove the same number of stones from all piles belonging to $k$ different groups, then because the piles originally from different groups still have different numbers of stones after the operation, these piles will still belong to $k$ different groups. The remaining untouched piles will still be in $n-k$ groups, of course. Therefore, after the operation, the number of different groups is at least $\max \{k, n - k\}$.

Thus, after each operation, the number of distinct groups $n$ does not decrease by more than half. This means that the number of groups after each operation is at least
$$
995,498,249,125,63,32,16 ; 8,4,2,1 .
$$

When only one pile is left, it is obviously the pile with 0 stones, meaning all the stones have been taken. This shows that at least 11 operations are required to complete the task.

In fact, if we sequentially take the numbers in the above sequence as $k$, keeping the piles with 0, 1, 2, ..., $k-1$ stones (with the first operation being an exception, where we keep the piles with 1, 2, ..., 995 stones) unchanged, and remove $k$ stones from each of the other piles. Clearly, this can be completed in just 11 operations. Therefore, the minimum number of operations required is 11.