8. Let $S=\{1,2, \cdots, 100\}$. If a three-element subset $A=\{a, b, c\}$ of $S$ satisfies $a+b=3 c$, then $A$ is said to have property $P$. The number of all three-element subsets of $S$ that have property $P$ is $\qquad$.
8. 1600.

Let $S_{k}=\{n \mid n \equiv k(\bmod 3), n \in S\}$, where $k=0,1,2$. Then
$$
\left|S_{0}\right|=\left|S_{2}\right|=33,\left|S_{1}\right|=34 \text {. }
$$

If $A=\{a, b, c\}$ has property $P$, then $a, b \in S_{0}$, or one of $a, b$ belongs to $S_{1}$ and the other belongs to $S_{2}$.

When $a, b$ both belong to $S_{0}$ and $a \neq b$, the unordered pair $\{a, b\}$ has $\mathrm{C}_{33}^{2}$ pairs; when one of $a, b$ belongs to $S_{1}$ and the other belongs to $S_{2}$, the unordered pair $\{a, b\}$ has $\mathrm{C}_{33}^{1} \mathrm{C}_{34}^{1}$ pairs.

Therefore, the number of unordered pairs $\{a, b\}$ that satisfy $a+b \equiv 0(\bmod 3)$ is $\mathrm{C}_{33}^{2}+\mathrm{C}_{33}^{1} \mathrm{C}_{34}^{1}$.

However, among these, the different unordered pairs $\{a, b\}$ where $a=2 b$ or $b=2 a$ do not form a three-element set $A$ when $c=b$ or $c=a$. Such unordered pairs are $\{k, 2 k\}(k=1,2, \cdots, 50)$, totaling 50 pairs.
Thus, the number of three-element subsets with property $P$ is
$$
C_{33}^{2}+C_{33}^{1} C_{34}^{1}-50=1600
$$