1. From the numbers $1,2, \cdots, 2014$, what is the maximum number of numbers that can be selected such that none of the selected numbers is 19 times another?
According to the problem, if $k$ and $19 k$ cannot both appear in $1,2, \cdots, 2014$, and since $2014=19 \times 106$, and $106=5 \times 19+11$, then choose
$$
1,2,3,4,5,106,107, \cdots, 2013 \text {, }
$$

These 1913 numbers satisfy the requirement.
The numbers not chosen are $6,7, \cdots, 105,2014$, a total of 101 numbers.

Since $2014=1913+101$, if 1914 numbers are selected, then in the following 101 pairs of numbers
$$
(6,6 \times 19),(7,7 \times 19), \cdots,(106,106 \times 19)
$$
there must be a pair where both numbers are selected.
Thus, there must be one number that is 19 times the other.
Therefore, from the 2014 numbers $1,2, \cdots, 2014$, the maximum number of numbers that can be selected is 1913.