Example 9 Given a natural number $n \geqslant 2$, find the smallest positive number $\lambda$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, and any $n$ numbers $b_{1}, b_{2}, \cdots, b_{n}$ in $\left[0, \frac{1}{2}\right]$, if $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}=1$, then $a_{1} a_{2} \cdots a_{n} \leqslant \lambda\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right)$.
By Cauchy-Schwarz inequality, we have
\[ 1 = \sum_{i=1}^{n} b_{i} = \sum_{i=1}^{n} \frac{\sqrt{b_{i}}}{\sqrt{a_{i}}} \sqrt{a_{i} b_{i}} \leqslant \left( \sum_{i=1}^{n} \frac{b_{i}}{a_{i}} \right)^{\frac{1}{2}} \left( \sum_{i=1}^{n} a_{i} b_{i} \right)^{\frac{1}{2}}, \]
thus
\[ \frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} \frac{b_{i}}{a_{i}}. \]
Let \( M = a_{1} a_{2} \cdots a_{n}, A_{i} = \frac{M}{a_{i}}, i=1,2, \cdots, n \), then
\[ \frac{M}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} b_{i} A_{i}. \]
By the rearrangement inequality, without loss of generality, assume \( b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, A_{1} \geqslant A_{2} \geqslant \cdots \geqslant A_{n} \).
Thus,
\[ \sum_{i=1}^{n} b_{i} A_{i} \leqslant b_{1} A_{1} + (1 - b_{1}) A_{2}, \]
Since \( 0 \leqslant b_{1} \leqslant \frac{1}{2}, A_{1} \geqslant A_{2} \), we have
\[ \sum_{i=1}^{n} b_{i} A_{i} \leqslant \frac{1}{2} (A_{1} + A_{2}) = \frac{1}{2} (a_{1} + a_{2}) a_{3} \cdots a_{n}. \]
By the AM-GM inequality, noting that \( \sum_{i=1}^{n} a_{i} = 1 \), we get
\[ \sum_{i=1}^{n} b_{i} A_{i} \leqslant \frac{1}{2} \left( \frac{1}{n-1} \right)^{n-1}. \]
Therefore, the required \( \lambda \leqslant \frac{1}{2} \left( \frac{1}{n-1} \right)^{n-1} \).
On the other hand, when \( a_{1} = a_{2} = \frac{1}{2(n-1)}, a_{3} = \cdots = a_{n} = \frac{1}{n-1}, b_{1} = b_{2} = \frac{1}{2}, b_{3} = \cdots = b_{n} = 0 \), we have
\[ a_{1} a_{2} \cdots a_{n} = \frac{1}{2} \left( \frac{1}{n-1} \right)^{n-1} \sum_{i=1}^{n} a_{i} b_{i}, \]
so \( \lambda \geqslant \frac{1}{2} \left( \frac{1}{n-1} \right)^{n-1} \).
In conclusion, \( \lambda = \frac{1}{2} \left( \frac{1}{n-1} \right)^{n-1} \).