19. There are black and white chess pieces in the bag, the number of black pieces is twice the number of white pieces. Each time, 3 black pieces and 2 white pieces are taken out from the bag. After a certain number of times, 1 white piece and 31 black pieces are left. How many black pieces were originally in the bag?
【Analysis】Since the number of black stones is twice the number of white stones, we can assume that the number of black stones taken each time is also twice that of the white stones, meaning 2 times 2 = 4 black stones and 2 white stones each time. When there is 1 white stone left, there will be 2 black stones left. Now, each time 4 - 3 = 1 fewer black stones are taken, so the excess number of black stones, divided by the difference between the intended and actual number taken, gives the number of times taken. Solve accordingly.

【Solution】Assuming the number of black stones taken each time is also twice that of the white stones, i.e., 2 times 3 = 6 black stones and 3 white stones each time, then:
$$
\begin{array}{l}
(31-1 \times 2) \div(2 \times 2-3) \\
= 29 \div 1 \\
= 29 \text { (times) } \\
3 \times 29+31 \\
= 87+31 \\
= 118 \text { (pieces) }
\end{array}
$$

Answer: There were originally 118 black stones in the bag.
The answer is: 118.
【Comment】The key to this problem is to assume that the number of black stones taken each time is twice that of the white stones, based on the fact that the number of black stones is twice that of the white stones. Solve by considering the difference between the intended and actual number of black stones taken, and the difference between the intended and actual remaining black stones.