8. An ordered sequence of positive integers $\left(a_{1}, a_{2}, \cdots, a_{23}\right)$ satisfies:
(1) $a_{1}<a_{2}<\cdots<a_{23}=50$,
(2) any three numbers in the sequence can form the sides of a triangle.
Then the number of sequences that satisfy the conditions is
8. 2576.

From condition (2), we know $a_{1}+a_{2}>a_{23}$.
Also, $a_{1}+a_{2}>a_{23}$ implies that for any $1 \leqslant i < j < k \leqslant 23$,
$$
a_{i} + a_{j} > a_{k}
$$
$\Rightarrow$ Any three numbers in the array can form a triangle.
Thus, condition (2) $\Leftrightarrow a_{1}+a_{2}>a_{23}$.
From $2 a_{2}>a_{1}+a_{2}>a_{23}=50$
$$
\Rightarrow a_{2}>25 \Rightarrow a_{2} \geqslant 26 \text{.}
$$

From $50=a_{23} \geqslant a_{22}+1 \geqslant a_{21}+2 \geqslant \cdots$
$$
\begin{aligned}
\geqslant & a_{2}+21 \\
\Rightarrow & a_{2} \leqslant 29 .
\end{aligned}
$$

Let $M=\left(a_{3}, a_{4}, \cdots, a_{21}, a_{22}\right)$.
From $a_{2}>a_{1}>50-a_{2}$
$$
\Rightarrow a_{2}-1 \geqslant a_{1} \geqslant 51-a_{2} \text{.}
$$

Thus, when $a_{2}=29$, $a_{1}=22,23, \cdots, 28$, $M$ has
$\mathrm{C}_{20}^{20}$ ways to choose;
When $a_{2}=28$, $a_{1}=23,24, \cdots, 27$, $M$ has $\mathrm{C}_{21}^{20}$ ways to choose;
When $a_{2}=27$, $a_{1}=24,25,26$, $M$ has $\mathrm{C}_{22}^{20}$ ways to choose; When $a_{2}=26$, $a_{1}=25$, $M$ has $\mathrm{C}_{23}^{20}$ ways to choose. Therefore, the total number of arrays that satisfy the conditions is
$$
7 \mathrm{C}_{20}^{20}+5 \mathrm{C}_{21}^{20}+3 \mathrm{C}_{22}^{20}+\mathrm{C}_{23}^{20}=2576 .
$$