8. In the set of the first ten thousand positive integers $\{1,2$, $\cdots, 10000\}$, the elements that leave a remainder of 2 when divided by 3, a remainder of 3 when divided by 5, and a remainder of 4 when divided by 7 are $\qquad$ in number.
8. 95 .

From the problem, we know that for each number $n$ that satisfies the condition, the number $2n$ should leave a remainder of 1 when divided by $3$, $5$, and $7$, and it should be even.

Therefore, $2n-1$ should be a common multiple of $3$, $5$, and $7$, and it should be odd, i.e., $2n-1$ should be an odd multiple of 105.

When $n \in \{1,2, \cdots, 10000\}$, $2n-1 \in$ $\{1,2, \cdots, 19999\}$, and among these, there are 190 numbers that are multiples of 105, of which exactly 95 are odd multiples.