67. By Cauchy's inequality,
$$\begin{array}{l} \left(\frac{b_{1}^{3}}{a_{1}}+\frac{b_{2}^{3}}{a_{2}}+\cdots+\frac{b_{n}^{3}}{a_{n}}\right)\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right) \geqslant \\ \left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}\right)^{2}= \\ \sqrt{\left(b_{1}^{2}+b_{2}^{2}++b_{n}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}++b_{n}^{2}\right)^{3}}= \\ \sqrt{\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}\right)\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)} \geqslant \\ a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \end{array}$$

Since $a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}>0$, we have $\frac{b_{1}^{3}}{a_{1}}+\frac{b_{2}^{3}}{a_{2}}+\cdots+\frac{b_{n}^{3}}{a_{n}} \geqslant 1$.