7.4. There are three piles of stones, and it is allowed to add stones to any pile, with the number of stones added must equal the sum of the number of stones in the other two piles at that time; it is also possible to remove stones from any pile when possible, with the number of stones removed equal to the sum of the number of stones in the other two piles at that time (for example, if there are 4, 7, and 12 stones in the three piles, then 19 stones can be added to the pile with 4 stones; or 11 stones can be removed from the pile with 12 stones). Now, suppose the three piles originally have 1993, 199, and 18 stones respectively. Can it be done through several operations to make one of the piles empty?
7.4. Cannot.

Let the method of adding stones be $A$, and the method of reducing stones be $B$. For any given three piles of stones, let the $i$-th operation be $c_{i}(i=1,2, \cdots)$. To obtain the pile with the fewest stones, since $A$ operations are limited, there must be a case where “$c_{i}=A, c_{i+1}=B$”. In this case, the $A$ and $B$ operations cancel each other out. Repeating this process, only $B$ operations remain, which means
$$
\begin{array}{l}
(1993,199,18) \rightarrow(1993-199-18,199,18) \rightarrow \cdots \\
\rightarrow(1993-(199+18) \times 9,199,18)=(40,199,18) \\
\rightarrow \cdots \rightarrow(40,25,18) .
\end{array}
$$

Thus, the fewest number of stones is 18.