8. Given a positive integer $n \geqslant 3$. Find the smallest real number $k$, such that for any positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\sum_{i=1}^{n-1} \frac{a_{i}}{s-a_{i}}+\frac{k a_{n}}{s-a_{n}} \geqslant \frac{n-1}{n-2},
$$

where, $s=a_{1}+a_{2}+\cdots+a_{n}$. (Proposed by Bin Zhao)
8. $k_{\text {min }}=\left(\frac{n-1}{n-2}\right)^{2}$.

On one hand, let
$$
a_{1}=a_{2}=\cdots=a_{n-1}=1, a_{n}=x>0 .
$$

Then, $\frac{n-1}{n-2+x}+\frac{k x}{n-1} \geqslant \frac{n-1}{n-2}$.
Thus, for any $x>0$, we have
$$
k \geqslant \frac{(n-1)^{2}}{(n-2)(n-2+x)} \text {. }
$$

Therefore, $k \geqslant\left(\frac{n-1}{n-2}\right)^{2}$.
On the other hand, we need to prove that the original inequality holds when $k=\left(\frac{n-1}{n-2}\right)^{2}$.
$$
\text { Let } \sum_{i=1}^{n-1} \frac{a_{i}}{s-a_{i}}+\frac{k a_{n}}{s-a_{n}}=A \text {. }
$$

By the Cauchy-Schwarz inequality, we know
$$
\begin{array}{l}
A \sum_{i=1}^{n} a_{i}\left(s-a_{i}\right) \geqslant\left(\sum_{i=1}^{n} a_{i}+\frac{n-1}{n-2} a_{n}\right)^{2} \\
\Rightarrow A \geqslant \frac{\left(t+\frac{n-1}{n-2} a_{n}\right)^{2}}{\left(t+a_{n}\right)^{2}-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}},
\end{array}
$$

where $t=a_{1}+a_{2}+\cdots+a_{n-1}$.
By the Cauchy-Schwarz inequality again, we get
$$
\begin{array}{l} 
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n-1}^{2} \geqslant \frac{t^{2}}{n-1} \\
\Rightarrow A \geqslant \frac{\left(t+\frac{n-1}{n-2} a_{n}\right)^{2}}{\left(t+a_{n}\right)^{2}-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}} \\
\geqslant \frac{\left(t+\frac{n-1}{n-2} a_{n}\right)^{2}}{\left(t+a_{n}\right)^{2}-\frac{t^{2}}{n-1}-a_{n}^{2}} \\
\geqslant \frac{t^{2}+2 \cdot \frac{n-1}{n-2} t a_{n}}{\frac{n-2}{n-1} t^{2}+2 t a_{n}}=\frac{n-1}{n-2} .
\end{array}
$$