8. Find the largest real number $a$ such that for all positive integers $n$ and all real numbers $x_{0}, x_{1}, \cdots, x_{n}\left(0=x_{0}<x_{1}<\cdots\right.$ $\left.<x_{n}\right)$, we have
$$
\sum_{i=1}^{n} \frac{1}{x_{i}-x_{i-1}} \geqslant a \sum_{i=1}^{n} \frac{i+1}{x_{i}} .
$$
8. The largest real number $a=\frac{4}{9}$.

If $a=\frac{4}{9}$, for each $2 \leqslant k \leqslant n$, by the Cauchy-Schwarz inequality we get
$$
\begin{array}{l}
\left(x_{k-1}+\left(x_{k}-x_{k-1}\right)\right)\left(\frac{(k-1)^{2}}{x_{k-1}}+\frac{3^{2}}{x_{k}-x_{k-1}}\right) \\
\geqslant(k-1+3)^{2} \\
\Rightarrow \frac{9}{x_{k}-x_{k-1}} \geqslant \frac{(k+2)^{2}}{x_{k}}-\frac{(k-1)^{2}}{x_{k-1}} .
\end{array}
$$

For $k=2,3, \cdots, n$, summing up the above inequality and adding $\frac{9}{x_{1}}$ to both sides, we get
$$
\begin{array}{l}
9 \sum_{k=1}^{n} \frac{1}{x_{k}-x_{k-1}} \geqslant 4 \sum_{k=1}^{n} \frac{k+1}{x_{k}}+\frac{n^{2}}{x_{n}} \\
>4 \sum_{k=1}^{n} \frac{k+1}{x_{k}} .
\end{array}
$$

This shows that when $a=\frac{4}{9}$, inequality (1) holds.
Define the sequence
$$
x_{0}=0, x_{k}=x_{k-1}+k(k+1)\left(k \in \mathbf{Z}_{+}\right) \text {. }
$$

Then $x_{k}=\sum_{i=1}^{k} i(i+1)=\frac{k(k+1)(k+2)}{3}$.
Thus, the left side of inequality (1)
$$
\begin{array}{l}
=\sum_{k=1}^{n} \frac{1}{k(k+1)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right) \\
=1-\frac{1}{n+1} ;
\end{array}
$$

The right side of inequality (1)
$$
\begin{array}{l}
=a \sum_{k=1}^{n} \frac{k+1}{\frac{k(k+1)(k+2)}{3}} \\
=a \sum_{k=1}^{n} \frac{3}{k(k+2)} \\
=3 a \sum_{k=1}^{n} \frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right) \\
=\frac{3}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right) a .
\end{array}
$$

When $n \rightarrow+\infty$,
the left side of inequality (1) $\rightarrow 1$, the right side $\rightarrow \frac{9}{4} a$.
Therefore, the maximum value of $a$ is $\frac{4}{9}$.