19. By Cauchy-Schwarz inequality,
$$1=\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n}\left(\frac{\sqrt{b_{i}}}{\sqrt{a_{i}}} \cdot \sqrt{a_{i} b_{i}}\right) \leqslant\left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{\frac{1}{2}}\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{\frac{1}{2}}$$

Thus,
$$\frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} \frac{b_{i}}{a_{i}}$$

Let \( M=a_{1} a_{2} \cdots a_{n}, A=\frac{M}{a_{i}}, i=1,2, \cdots, n \), then \(\frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} b_{i} A_{i}\).
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 we have
$$\sum_{i=1}^{n} b_{i} A_{i} \leqslant b_{1} A_{1}+\left(1-b_{1}\right) A_{2}$$

Since \( 0 \leqslant b_{1} \leqslant \frac{1}{2}, A_{1} \geqslant A_{2} \), we have
$$\sum_{i=1}^{n} b_{i} A_{i} \leqslant \frac{1}{2}\left(A_{1}+A_{2}\right)=\frac{1}{2}\left(a_{1}+a_{2}\right) a_{3} \cdots a_{n}$$

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, \(\lambda \leqslant \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}\).
On the other hand, when \( a_{1}=a_{2}=\frac{1}{2(n-1)}, a_{3}=\cdots=a_{n}=\frac{1}{n-1}, b_{1}=b_{2}=\frac{1}{2} \), \( b_{3}=\cdots=b_{n}=0 \),
$$a_{1} a_{2} \cdots a_{n}=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1} \sum_{i=1}^{n} a_{i} b_{i}$$

Thus, \(\lambda \geqslant \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}\).
In conclusion, \(\lambda=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}\).
19. Given a positive integer $n \geqslant 2$. Find the smallest positive number $\lambda$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ and any $n$ positive numbers $b_{1}, b_{2}, \cdots, b_{n}$ in $[0, \left.\frac{1}{2}\right]$, if $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}=1$, then $a_{1} a_{2} \cdots a_{n} \leqslant \lambda\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right)$. (1992 China National Training Team Selection Exam Problem)