1. First, construct the following 1009 fractions:
$$
\frac{2}{1011}, \frac{3}{1012}, \cdots, \frac{1010}{2019} \text{, }
$$

Among them, the largest number is $\frac{1010}{2019}$.
Next, we prove that this is the minimum value.
Proof by contradiction.
Assume there are another 1009 fractions that meet the conditions, and the largest number is less than $\frac{1010}{2019}$.

Since 2019 cannot be a numerator, otherwise the largest number would be greater than 1, we can denote the numerator of the fraction with 2019 as the denominator as $a$.
Clearly, $a < \frac{1010}{2019}$,
$$

Contradiction.
1. Use the numbers $2,3, \cdots, 2019$ to form 1009 fractions, with each number appearing only once in either the numerator or the denominator. From these 1009 fractions, select the largest one. Find the minimum value of this number.