Problem 842

9. Since $1990=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+0 \cdot 2^{5}+0 \cdot 2^{4}+0 \cdot 2^{3}+2^{2}+2^{1}+0 \cdot 2^{0}$, and writing $1,2, \cdots$, 1989 in binary form, the operation is as follows:

The first time, take away $2^{10}=1024$ stones from each pile that has enough; the second time, take away $2^{9}=512$ stones from each pile that has enough, $\cdots$, finally, take away the stones from the piles that have only $2^{0}=1$ stone left. In this way, a total of 11 times are used, because the pile with one stone must have an operation to take one stone. If the remaining operations take away more than 2 stones each time, then the pile with exactly two stones cannot be taken away. Therefore, 2 operations can take at most $1+2=3$ stones, $\cdots$, 10 operations can take at most $1+2+2^{2}+\cdots+2^{9}=1023$ stones, so 1990 piles of stones must undergo at least 11 operations.