8. Let the set $S=\{1,2, \cdots, 100\}$, for a positive integer $m$, in any $m$-element subset of set $S$, there must be one number that is a divisor of the product of the other $m-1$ numbers. Then the smallest possible value of $m$ is $\qquad$ .
8. 26 .

All prime numbers not greater than 100 total 25, forming the set
$$
\begin{aligned}
T= & \{2,3,5,7,11,13,17,19,23,29,31, \\
& 37,41,43,47,53,59,61,67,71,73, \\
& 79,83,89,97\} .
\end{aligned}
$$

Notice that, each element in set $T$ cannot be divided by the product of the other 24 elements.
Thus, $m > |T| = 25 \Rightarrow m \geqslant 26$.
On the other hand, we prove by contradiction: for any 26-element subset of set $S$, there must be a number that is a divisor of the product of the other 25 numbers.

For convenience, for a prime number $p$ and a positive integer $x$, let $\alpha_{p}(x)$ denote the exponent of $p$ in the prime factorization of $x$.

If there exists a 26-element subset $A$ of set $S$ such that for each $x \in A$, $x$ does not divide the product of the other 25 numbers in $A$, then for each $x \in A$, there exists a prime factor $p$ of $x$ such that $\alpha_{p}(x) > \alpha_{p}\left(\prod_{x \in A \backslash \{x\}} z\right)$, and we call such a prime $p$ a special prime factor of $x$. Such special prime factors are not necessarily unique.

Since $|A| = 26$, and all special prime factors belong to set $S$, and set $S$ contains only 25 prime numbers, there must be two different elements $x$ and $y$ in set $A$ that have the same special prime factor $p$.
By the special property and $y \in A \backslash \{x\}$, we have
$$
\begin{array}{l}
\alpha_{p}(x) > \alpha_{p}\left(\prod_{z \in A \backslash \{x\}} z\right) \geqslant \alpha_{p}(y). \\
\text{Similarly, } \alpha_{p}(y) > \alpha_{p}\left(\prod_{z \in A \backslash \{y\}} z\right) \geqslant \alpha_{p}(x),
\end{array}
$$

which is a contradiction.
In conclusion, the smallest possible value of $m$ is 26.