9. There are four teacups with their mouths facing up. Now, each time three of them are flipped, and the flipped teacups are allowed to be flipped again. After $n$ flips, all the cup mouths are facing down. Then the minimum value of the positive integer $n$ is $\qquad$ .
9.4 .

Let $x_{i}$ be the number of times the $i$-th cup ($i=1,2,3,4$) is flipped when all cup mouths are facing down, then $x_{i}$ is odd.
From $x_{1}+x_{2}+x_{3}+x_{4}=3 n$, we know that $n$ is even.
It is easy to see that when $n=2$, the condition is not satisfied, hence $n \geqslant 4$.
When $n=4$, use 1 to represent the cup mouth facing up, and 0 to represent the cup mouth facing down. The following flips can satisfy the condition:
$$
\begin{array}{l}
(1,1,1,1) \xrightarrow{1}(0,0,0,1) \xrightarrow{2}(0,1,1,0) \\
\xrightarrow{3}(1,0,1,1) \xrightarrow{4}(0,0,0,0) .
\end{array}
$$

Therefore, the minimum value of the positive integer $n$ is 4.