4. Let the set $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$, satisfying $(a+b) \mid a b$.
First, by enumeration, we obtain 23 pairs $(a, b)$, each of which satisfies $(a+b) \mid a b$.

Construct a 50-order graph $G$ (with the set $S$ of numbers $1,2, \cdots, 50$ as vertices, and if two numbers $a, b$ belong to the above pairs, then let $a, b$ be adjacent). Thus, the graph $G$ has exactly 23 edges (isolated points in Figure 11 are not labeled).

Remove some points so that the edges are also removed. At least 12 points need to be removed, for example, removing the points in the set
$$
M=\{3,5,7,9,10,12,14,16,21,24,30,36\}
$$

At this point, the graph $G$ has $50-12=38$ points left, and these 38 numbers form the set $A$, which does not contain any of the above pairs, i.e., no pair in $A$ satisfies the condition.
Therefore, the smallest positive integer $k \geqslant 39$.
Next, we prove that $k=39$ satisfies the condition.
First, take 12 edges from the graph $G$ that have no common vertices (12 pairs of numbers):
$$
\begin{array}{l}
(9,18),(36,45),(5,20),(15,30), \\
(10,40),(8,24),(16,48),(4,12), \\
(3,6),(7,42),(21,28),(14,35),
\end{array}
$$

Each edge has exactly one vertex in the set $M$.
These 12 edges have 24 vertices in total, and the 24 numbers form a 24-element subset $B$ of the set $S$; the remaining 26 numbers in the set $S$ form the subset $C$, i.e.,
$$
S=B \cup C, B \cap C=\varnothing \text {. }
$$

Now, take any 39-element subset $T$ from the set $S$, then $T$ can have at most 26 numbers from the set $C$, which means it must have at least 13 numbers from the set $B$, and among these, there must be two numbers from the same pair of the above 12 pairs, say $(a, b)$, then $(a+b) \mid a b$.
Therefore, the smallest $k$ that satisfies the condition is 39.