4. Let 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.
(《Mathematics in Middle School》1999 Issue 1 Olympiad Problems)
4. The required $\max a=665$. First, we prove: $a \leqslant 665$. Clearly, the 999-element subset $X_{0}=\{999,1000,1001, \cdots, 1997 \mid$ does not contain $x, y \in X_{0}$ such that $x \cdot 1997$, which is greater than the largest element in $X_{0}$, holds. Thus, $a$ cannot be any of the numbers $999,1000,1001, \cdots, 1997$.
Next, we prove that $a \neq 666,667,668, \cdots, 997,998$. Construct the sets:
$B_{i}=\{666+i\} \cup X_{0} \backslash\{2(666+i)\}, i=0,1, \cdots, 332$.
For $B_{i}$, $(666+i) \cdot 3 \geqslant 1998$, and $(666+i) \cdot 2 \notin B_{i}$, so no other multiples of $666+i$ are in $B_{i}$. It has already been 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 $666+i(i=0,1, \cdots, 332)$. Thus, the multiples (excluding itself) of any other elements in $B_{i}$ are not in $B_{i}$. Therefore, there do not exist two elements in $B_{i}$ such that $x<y, x \mid y$, and $B_{i}$ $(i=0,1, \cdots, 332)$ includes 666, $667, \cdots, 998$. Hence, $a \neq 666,667, \cdots, 998$, so $a \leqslant 665$.

Next, we prove that 665 is attainable. Assume 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 $665 \cdot 2,665 \cdot 3 \notin X$. Construct the following 997 drawers, which include all elements of $A$ except $665,665 \cdot 2,665 \cdot 3$, and each element appears once: $K_{1}=\left|1,1 \cdot 2,1 \cdot 2^{2}, \cdots, 1 \cdot 2^{10}\right|, K_{2}=\left|3,3 \cdot 2,3 \cdot 2^{2}, \cdots, 3 \cdot 2^{9}\right|, K_{3}=\mid 5,5 \cdot 2$, $\left.5 \cdot 2^{2}, \cdots, 5 \cdot 2^{8}\right\}, K_{4}=\left\{7,7 \cdot 2,7 \cdot 2^{2}, \cdots, 7 \cdot 2^{8} \mid, \cdots, K_{332}=\{663,663 \cdot 2,663 \cdot 3\}, K_{334}=\{667,667 \cdot\right.$ $2 \mid, K_{335}=\{669,669 \cdot 2\}, \cdots, K_{995}=\{1991\}, K_{996}=\{1993\}, K_{997}=\{1997\}$. Thus, the other 998 elements of $X$ (excluding 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, leading to a contradiction. Proof complete.