9. If the positive integer $m$ makes it true that for any set of positive numbers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$

then the minimum value of the positive integer $m$ is $\qquad$
9.3.

Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$

It is verified that $m=1, m=2$ do not meet the requirements.
Hence $m \geqslant 3$.
$$
\begin{array}{l}
\text { By } \frac{a_{1}^{3}+a_{2}^{3}+a_{3}^{3}}{3} \geqslant a_{1} a_{2} a_{3}, \\
\frac{a_{1}^{3}+a_{2}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{2} a_{4}, \\
\frac{a_{1}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{3} a_{4}, \\
\frac{a_{2}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{2} a_{3} a_{4}, \\
a_{1} a_{2} a_{3} a_{4}=1, \\
\text { then } a_{1}^{3}+a_{2}^{3}+a_{3}^{3}+a_{4}^{3} \\
\geqslant a_{1} a_{2} a_{3}+a_{1} a_{2} a_{4}+a_{1} a_{3} a_{4}+a_{2} a_{3} a_{4} \\
=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}} .
\end{array}
$$

Therefore, $m=3$ meets the requirements:
Thus, the smallest positive integer $m$ is 3.