To find the smallest positive integer $K$ such that every $K$-element subset of $\{1, 2, \ldots, 50\}$ contains two distinct elements $a$ and $b$ such that $a + b$ divides $ab$, we need to analyze the properties of the set and the divisibility condition.

Consider the set $\{1, 2, \ldots, 50\}$. We need to ensure that in any subset of size $K$, there exist two elements $a$ and $b$ such that $a + b \mid ab$.

First, observe that if $a$ and $b$ are both even, then $a + b$ is even and $ab$ is even, so $a + b$ divides $ab$. Similarly, if $a$ and $b$ are both odd, then $a + b$ is even and $ab$ is odd, so $a + b$ does not necessarily divide $ab$.

To ensure that $a + b \mid ab$ for any subset of size $K$, we need to consider the worst-case scenario where no two elements $a$ and $b$ satisfy the condition. This happens when the subset contains numbers that do not pair well under the given condition.

By the Pigeonhole Principle, if we have more than 25 elements in the subset, there must be at least one pair of elements $a$ and $b$ such that $a + b$ divides $ab$, because there are only 25 possible sums $a + b$ that are less than or equal to 50.

Therefore, the smallest positive integer $K$ such that every $K$-element subset of $\{1, 2, \ldots, 50\}$ contains two distinct elements $a$ and $b$ such that $a + b$ divides $ab$ is:

The answer is: $\boxed{26}$.