19. (6 points) There are 2015 lamps numbered $1, 2, 3, \cdots 2015$, each controlled by a pull switch. If the pull switches of lamps numbered as multiples of 2, 3, and 5 are each pulled once, how many lamps will still be on? 

So, the number of lamps that are on is: $2015-1009=1006$ (lamps).
Answer: At this time, there are 1006 lamps on.
Therefore, the answer is: 1006.
【Solution】Solution: Among the 2015 numbers from 1 to 2015,
Multiples of 2: $2015 \div 2 \approx 1007$ (numbers)
Multiples of 3: $2015 \div 3 \approx 671$ (numbers)
Multiples of 5: $2015 \div 5=403$ (numbers)
Multiples of 2 and 3: $2015 \div(2 \times 3) \approx 335$ (numbers)
Multiples of 2 and 5: $2015 \div(2 \times 5) \approx 201$ (numbers)
Multiples of 3 and 5: $2015 \div(3 \times 5)=\approx 134$ (numbers)
Multiples of 2, 3, and 5: $2015 \div(2 \times 3 \times 5) \approx 67$ (numbers)
It is known that, the number of lamps pulled three times: 67,
The number of lamps pulled twice: $(335-67)+(201-67)+(134-67)=268+134+67=469$ (lamps)
The number of lamps pulled once:
$$
\begin{aligned}
& (1007-268-134-67)+(671-268-67-67)+(403-134-67-67) \\
= & 538+269+135 \\
= & 942 \text { (lamps) }
\end{aligned}
$$

The number of lamps turned off: $942+67=1009$ (lamps)