74. Let the set $A=\{1,2,3, \cdots, 1997\}$, for any 999-element subset $X$ of $A$, if there exist $x, y \in X$, such that $x<y$ and $x \mid y$, then $X$ is called a good set. Find the largest natural number $a(a \in A)$, such that any 999-element subset containing $a$ is a good set.
Solution: We prove that $\max a=665$.
First, we prove that $a \leqslant 665$. Clearly, the 999-element subset $X_{0}$ $=\{999,1000,1001, \cdots, 1997\}$ does not contain any $x, y \in$ $X_{0}$ such that $x1997$, i.e., larger than the maximum element of $X_{0}$, which holds. Thus, $a$ cannot be any of the numbers $999,1000,1001, \cdots$, 1997. 998.
Next, we prove that $2 \neq 666$ construct
$0, \frac{1}{x} \frac{1}{52}$.
For $B_{i}$, $(666+i) \times 3 \geqslant 1998$, and $(666$ $+i) \times 2 \bar{\epsilon} B_{i}$, so no other multiples of $666+i$ are in $B_{i}$.

We have already proven that no non-multiples of any element in $X_{0}$ are in $X_{0}$, and $666 + i < 999 (i=0,1,2, \cdots, 332)$, so no multiples of any element in $X_{0}$ can be $666+i (i=0,1, \cdots, 332)$. This ensures that no multiples (other than itself) of any element in $B_{i}$ are in $B_{i}$. Thus, there do not exist two elements in $B_{i}$ such that $x<y$ and $x \mid y$. Since $B_{i}$ $(i=0,1, \cdots, 332)$ includes $666,667, \cdots$, 998, $a \neq 666,667, \cdots, 998$. Therefore, $a \geqslant 665$.

Next, we prove that 665 is achievable. Suppose there exists a 999-element subset $X$ containing 665, such that there do not exist $x, y \in X, x<y$ with $x \mid y$, then $665 \times 2,665 \times 3 \in X$.

Construct the following 997 drawers, which include all elements of $A$ except $665,665 \times 2,665 \times 3$, and each element appears only once:
$$
\begin{array}{l}
\left\{1,1 \times 2,1 \times 2^{2}, \cdots, 1 \times 2^{10}\right\} \\
\left\{3,3 \times 2,3 \times 2^{2}, \cdots, 3 \times 2^{9}\right\} \\
\left\{5,5 \times 2,5 \times 2^{2}, \cdots, 5 \times 2^{8}\right\} \\
\cdots \cdots \\
\{663,663 \times 2,663 \times 3\} \\
\{667,667 \times 2\} \\
\{669,669 \times 2\} \\
\cdots \cdots \cdots
\end{array}
$$
\{1991\} \{1993\} \{1997\}
The 998 elements of $X$ other than 665 are placed into these 997 drawers, so there must be two elements in the same drawer, and the elements in the same drawer are multiples of each other, which is a contradiction. Proof complete.
(Xiao Mingyu, Class 981, Department of Computational Mathematics and Applied Software, Central South University of Technology, 410012)