$35 \cdot 24$ boxes originally contained 7 small balls. A magician took out several balls from it, turned each ball into 7 balls, and put them back into the box. Then, he took out some balls again, turned each ball into 7 balls, and put them back into the box. This process continued, and at a certain moment, when the magician stopped taking balls and performing magic, the total number of balls in the box could be one of the following:
(A) 1990.
(B) 1991.
(C) 1992.
(D) 1993.
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1991)
[ Sol] From the problem, the magician always turns one ball into 7 balls each time, which means no matter how many balls are transformed each time, the total number of balls after transformation always increases by a multiple of 6.
Thus, regardless of how many transformations occur, the remainder when the total number of balls is divided by 6 remains unchanged.
Also, $7 \div 6=1 \cdots \cdots 1$, meaning the initial number of balls leaves a remainder of 1 when divided by 6.
And
$$
\begin{array}{ll}
1990 \div 6=331 \cdots \cdots 4, & 1991 \div 6=331 \cdots \cdots 5, \\
1992 \div 6=332, & 1993 \div 6=332 \cdots \cdots 1,
\end{array}
$$

Therefore, the total number of balls at some point can only be 1993.
Hence, the answer is $(D)$.