8. Given $x_{i} \in\{-1,1\}, i=1,2, \cdots, 2021$, and $x_{1}+x_{2}+\cdots+x_{k} \geqslant 0(k=1,2$, $\cdots, 2020), x_{1}+x_{2}+\cdots+x_{2021}=-1$. Then the number of ordered tuples $\left(x_{1}, x_{2}, \cdots, x_{2021}\right)$ is $\qquad$.
The Catalan number $C_{n}$ is the number of sequences $x_{1}, x_{2}, \cdots, x_{2 n}$ composed of $n$ 1s and $n$ -1s, which satisfy the condition $x_{1}+x_{2}+\cdots+x_{k} \geqslant 0, k=1,2, \cdots, 2 n$. In this sequence, at any position from left to right and all positions before it, the number of 1s will not be less than the number of -1s. Let $S_{n}$ be the set of sequences composed of $n$ 1s and $n$ -1s, and it is clear that $\left|S_{n}\right|=C_{2 n}^{n}$, which is the number of ways to determine the positions of 1s in $2 n$ positions.

Let $A_{n}$ be the set of sequences in $S_{n}$ that do not satisfy the condition. Then for any $\sigma=x_{1} x_{2} \cdots x_{2 n} \in A_{n}$, there must exist at least one $k$ for which the inequality does not hold. Let the first (smallest) such $k$ be $k_{0}$, then $k_{0}$ must be odd, and $x_{1}+x_{2}+\cdots+x_{k_{0}-1}=0, x_{k_{0}}=-1$.
By changing the signs of the first $k_{0}$ numbers, we get $\bar{\sigma}=\left(-x_{1}\right)\left(-x_{2}\right) \cdots\left(-x_{k_{0}}\right) x_{k_{0}+1} \cdots x_{2 n}$, which is a sequence composed of $n+1$ 1s and $n-1$ -1s. Conversely, any sequence composed of $n+1$ 1s and $n-1$ -1s must be the result of the above transformation applied to some sequence in $A_{n}$, because there always exists a value $k$ such that $x_{1} x_{2} \cdots x_{k}$ contains one more 1 than -1. Thus, the number of sequences in $A_{n}$ is equal to the number of sequences composed of $n+1$ 1s and $n-1$ -1s, i.e., $A_{n}=C_{2 n}^{n+1}$. Therefore, the number of sequences composed of $n$ 1s and $n$ -1s that satisfy the condition is $C_{n}=\left|S_{n}\right|-\left|A_{n}\right|=$ $C_{2 n}^{n}-C_{2 n}^{n+1}=\frac{1}{n+1} C_{2 n}^{n}$. In summary, it is clear that $x_{2021}=-1$ in this problem, so the number of ordered tuples $\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{2021}\right)$ is $C_{1010}=\frac{1}{1011} C_{2020}^{1010}$.