Eight, given that $n$ is a positive integer, and the real number $x$ satisfies
$$
|1-| 2-\cdots|(n-1)-| n-x|| \cdots \|=x \text {. }
$$

Find the value of $x$.
(Zhang Limin, problem contributor)
Conclusion 1 If $a \geqslant 0, b \geqslant 0$, then
$$
|a-b| \leqslant \max \{a, b\} \text {. }
$$

Conclusion 2 If $0 \leqslant x < k-1$, then
$$
\begin{array}{l}
|(k-1)-x|=x-(k-1) \\
<k-(k-1)=1 ;
\end{array}
$$

If $x \leqslant k-1$, then by Conclusion 1 we have
$$
|(k-1)-x| \leqslant k-1 \text {. }
$$

In summary, Conclusion 2 holds.
Conclusion 3 If $0 \leqslant x \leqslant 1, k \in \mathbf{N}_{+}$, then
$$
|k-| k+1-|k+2-| k+3-x||||=x \text {. }
$$

In fact, removing the absolute values from the left side of the above equation yields the validity of Conclusion 3.
Returning to the original problem.
By Conclusion 1, if $x \geqslant n$, then
$$
\begin{array}{l}
|1-| 2-\cdots-|n-x| \cdots|| \\
\leqslant \max \{1,|2-| 3-\cdots|n-x| \cdots||\} \\
\leqslant \cdots \\
\leqslant \max \{1,2, \cdots, n-1, x-n\}<x,
\end{array}
$$

which contradicts the given condition.
Therefore, $x < n$.
By Conclusion 2 and $|x-n| \leqslant n$, we get
$$
\begin{array}{l}
|(n-1)-| n-x|| \leqslant n-1, \\
|(n-2)-|(n-1)-|n-x||| \leqslant n-2, \\
\cdots \cdots
\end{array}
$$

Continuing in this manner, we obtain
$$
|1-| 2-\cdots-|n-x| \cdots \| \leqslant 1 \text {. }
$$

Thus, $0 \leqslant x \leqslant 1$.
If $n=4 k+1$ or $n=4 k+2(k \in \mathbf{N})$, by Conclusion 3, the given equation simplifies to
$$
|1-x|=x \Rightarrow x=\frac{1}{2} \text {. }
$$

If $n=4 k+3$ or $n=4 k(k \in \mathrm{N})$, by Conclusion 3, the given equation simplifies to
$$
|x|=x \text {. }
$$

Since $x \leqslant 1$, it follows that $0 \leqslant x \leqslant 1$.