1. There are 2019 bags, each containing 2019 beads with a total weight of 1 kilogram. In each bag, the beads are numbered as $1,2, \cdots, 2019$. A selection is called a proper selection if and only if beads with different numbers are chosen from different bags (at most one bead from each bag), and the total weight is at least 1 kilogram. Find the maximum value of $k$ such that there always exist at least $k$ different proper selections.
1.2018!.

A selection of beads with different numbers, one from each of 2019 bags, is called "good". Let $\left(\sigma_{1}, \sigma_{2}, \cdots, \sigma_{2019}\right)$ be a good selection, where $\sigma_{i}$ is the number of the bead selected from the $i$-th bag $(i=1,2, \cdots, 2019)$. Two good selections $\left(\sigma_{1}^{\prime}, \sigma_{2}^{\prime}, \cdots, \sigma_{2019}^{\prime}\right)$ and $\left(\sigma_{1}^{\prime \prime}, \sigma_{2}^{\prime \prime}, \cdots, \sigma_{2019}^{\prime \prime}\right)$ are called "equivalent" if and only if there exists an integer $m$ such that for each $i=1, 2, \cdots, 2019$,
$$
\sigma_{i}^{\prime}+m \equiv \sigma_{i}^{\prime \prime}(\bmod 2019),
$$

Each good selection is equivalent to exactly 2018 other good selections, and the set of all 2019! good selections can be divided into $\frac{2019!}{2019}=2018!$ equivalence classes.

Since any equivalence class contains 2019 selections that include each bead exactly once, the sum of the weights of all beads in this equivalence class is 2019. In each equivalence class, there is at least one good selection, so there are at least 2018! different good selections.

Below is an example where there are exactly 2018! good selections.
Assume the weights of the beads in the first bag are:
$$
\begin{array}{c}
\frac{1}{2019}+\frac{2018}{2019^{2}}, \frac{1}{2019}-\frac{1}{2019^{2}}, \frac{1}{2019}- \\
\frac{1}{2019^{2}}, \cdots, \frac{1}{2019}-\frac{1}{2019^{2}},
\end{array}
$$

The weights of the other beads are all $\frac{1}{2019}$.
Thus, each good selection must select the first bead from the first bag, otherwise, the total weight is at most $1-\frac{1}{2019^{2}}<1$; each good selection must include 2019 beads, otherwise, the total weight is at most
$$
\begin{array}{l}
\frac{1}{2019}+\frac{2018}{2019^{2}}+2017 \times \frac{1}{2019} \\
=1-\frac{1}{2019^{2}}<1 .
\end{array}
$$

Therefore, any good selection must include the first bead from the first bag and 2018 other beads, and there are 2018! such good selections.