6. Let $n$ be an integer greater than 1. The first $n$ prime numbers, in ascending order, are denoted as $p_{1}, p_{2}, \cdots, p_{n}$ (i.e., $p_{1}=2$, $p_{2}=3, \cdots$), and let $A=p_{1}^{p_{1}} p_{2}^{p_{2}} \cdots p_{n}^{p_{n}}$. Find all positive integers $x$ such that $\frac{A}{x}$ is even and $\frac{A}{x}$ has exactly $x$ distinct positive divisors.
6. From the given, we have $2 x \mid A$.

Notice that, $A=4 p_{2}^{p_{2}} p_{3}^{p_{3}} \cdots p_{n}^{p_{n}}$.
Therefore, we can set $x=2^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \cdots p_{n}^{\alpha_{n}}$, where $0 \leqslant \alpha_{1} \leqslant 1, 0 \leqslant \alpha_{i} \leqslant p_{i} (i=2,3, \cdots, n)$.
Then $\frac{A}{x}=2^{2-\alpha_{1}} p_{2}^{p_{2}-\alpha_{2}} p_{3}^{p_{3}-\alpha_{3}} \cdots p_{n}^{p_{n}-\alpha_{n}}$.
Thus, the number of different positive divisors of $\frac{A}{x}$ is
$$
\left(3-\alpha_{1}\right)\left(p_{2}-\alpha_{2}+1\right) \cdots\left(p_{n}-\alpha_{n}+1\right).
$$

From the given, we have
$$
\begin{array}{l}
\left(3-\alpha_{1}\right)\left(p_{2}-\alpha_{2}+1\right) \cdots\left(p_{n}-\alpha_{n}+1\right) \\
=x=2^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \cdots p_{n}^{\alpha_{n}}.
\end{array}
$$

Next, we use mathematical induction to prove: the array $\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right)$ that satisfies equation (1) must be $(1,1, \cdots, 1) (n \geqslant 2)$.
(1) When $n=2$, equation (1) becomes
$$
\left(3-\alpha_{1}\right)\left(4-\alpha_{2}\right)=2^{\alpha_{1}} \times 3^{\alpha_{2}} \left(\alpha_{1} \in\{0,1\}\right).
$$

If $\alpha_{1}=0$, then $3\left(4-\alpha_{2}\right)=3^{\alpha_{2}}$, there is no non-negative integer $\alpha_{2}$ that satisfies; if $\alpha_{1}=1$, then $2\left(4-\alpha_{2}\right)=2 \times 3^{\alpha_{2}}$, we get $\alpha_{2}=1$, thus, $\left(\alpha_{1}, \alpha_{2}\right)=(1,1)$, i.e., the conclusion holds when $n=2$.
(2) Assume that the conclusion holds when $n=k-1 (k \geqslant 3)$.
When $n=k$, equation (1) becomes
$$
\begin{aligned}
& \left(3-\alpha_{1}\right)\left(p_{2}-\alpha_{2}+1\right) \cdots \cdots \\
& \left(p_{k-1}-\alpha_{k-1}+1\right)\left(p_{k}-\alpha_{k}+1\right) \\
= & 2^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \cdots p_{k}^{\alpha_{k}}.
\end{aligned}
$$

If $\alpha_{k} \geqslant 2$, then considering
$$
\begin{array}{l}
0<p_{k}-\alpha_{k}+1<p_{k}, \\
0<p_{i}-\alpha_{i}+1 \leqslant p_{i}+1<p_{k} (1 \leqslant i \leqslant k-1).
\end{array}
$$

Thus, the left side of equation (2) cannot be divisible by $p_{k}$, but the right side of equation (2) is a multiple of $p_{k}$, which is a contradiction.
If $\alpha_{k}=0$, then equation (2) becomes
$$
\begin{array}{l}
\left(3-\alpha_{1}\right)\left(p_{2}-\alpha_{2}+1\right) \cdots \cdot \\
\left(p_{k-1}-\alpha_{k-1}+1\right)\left(p_{k}+1\right) \\
=2^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \cdots p_{k-1}^{\alpha_{k-1}}.
\end{array}
$$

Notice that, $p_{2}, p_{3}, \cdots, p_{k}$ are odd primes.
On one hand, $p_{k}+1$ is even, thus, the left side of equation (3) is even; on the other hand, the right side $p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \cdots p_{k-1}^{\alpha_{k-1}}$ is odd. Thus, there must be $\alpha_{1}=1$. But at this time, $3-\alpha_{1}=2$, so the left side is a multiple of 4, but the right side is not a multiple of 4, which is still a contradiction.
From the above discussion, we know that $\alpha_{k}=1$.
At this time, in equation (2), $p_{k}-\alpha_{k}+1=p_{k}^{\alpha_{k}}=p_{k}$.
Thus, $\left(3-\alpha_{1}\right)\left(p_{2}-\alpha_{2}+1\right) \cdots\left(p_{k-1}-\alpha_{k-1}+1\right)$ $=2^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \cdots p_{k-1}^{\alpha_{k-1}}$.
By the induction hypothesis, we know
$$
\alpha_{1}=\alpha_{2}=\cdots=\alpha_{k-1}=1.
$$

Thus, $\alpha_{1}=\alpha_{2}=\cdots=\alpha_{k-1}=\alpha_{k}=1$, i.e., the conclusion holds when $n=k$.
From (1) and (2), we know
$$
\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right)=(1,1, \cdots, 1).
$$

Therefore, the required positive integer is