Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$, we have  
$$ \frac{1}{x_{1}-x_{0}}+\frac{1}{x_{2}-x_{1}}+\cdots+\frac{1}{x_{n}-x_{n-1}} \geqslant a\left(\frac{2}{x_{1}}+\frac{3}{x_{2}}+\cdots+\frac{n+1}{x_{n}}\right) . $$
We first show that \( a = \frac{4}{9} \) is admissible. For each \( 2 \leqslant k \leqslant n \), by the Cauchy-Schwarz Inequality, we have
\[ \left(x_{k-1} + (x_{k} - x_{k-1})\right) \left(\frac{(k-1)^{2}}{x_{k-1}} + \frac{3^{2}}{x_{k} - x_{k-1}}\right) \geqslant (k-1+3)^{2}, \]
which can be rewritten as
\[ \frac{9}{x_{k} - x_{k-1}} \geqslant \frac{(k+2)^{2}}{x_{k}} - \frac{(k-1)^{2}}{x_{k-1}}. \]
Summing (2) over \( k = 2, 3, \ldots, n \) and adding \( \frac{9}{x_{1}} \) to both sides, we have
\[ 9 \sum_{k=1}^{n} \frac{1}{x_{k} - x_{k-1}} \geqslant 4 \sum_{k=1}^{n} \frac{k+1}{x_{k}} + \frac{n^{2}}{x_{n}} > 4 \sum_{k=1}^{n} \frac{k+1}{x_{k}}. \]
This shows (1) holds for \( a = \frac{4}{9} \). Next, we show that \( a = \frac{4}{9} \) is the optimal choice. Consider the sequence defined by \( x_{0} = 0 \) and \( x_{k} = x_{k-1} + k(k+1) \) for \( k \geqslant 1 \), that is, \( x_{k} = \frac{1}{3} k(k+1)(k+2) \). Then the left-hand side of (1) equals
\[ \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{n+1}, \]
while the right-hand side equals
\[ a \sum_{k=1}^{n} \frac{k+1}{x_{k}} = 3a \sum_{k=1}^{n} \frac{1}{k(k+2)} = \frac{3}{2} a \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+2} \right) = \frac{3}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) a. \]
When \( n \) tends to infinity, the left-hand side tends to 1 while the right-hand side tends to \( \frac{9}{4} a \). Therefore \( a \) has to be at most \( \frac{4}{9} \). Hence the largest value of \( a \) is \( \frac{4}{9} \).