$A, B, C, D$ four people attend a meeting, and they all receive the same positive integer. Next, each person describes this positive integer. Each person says three sentences, of which at least one is true and at least one is false. What they say is as follows:
$A:\left(A_{1}\right)$ This number is less than 12;
$\left(A_{2}\right) 7$ does not divide this number;
$\left(A_{3}\right) 5$ multiplied by this number is less than 70.
$B:\left(B_{1}\right) 12$ multiplied by this number is greater than 1000;
$\left(B_{2}\right) 10$ divides this number;
$\left(B_{3}\right)$ This number is greater than 100.
$C:\left(C_{1}\right) 4$ divides this number;
$\left(C_{2}\right) 11$ multiplied by this number is less than 1000;
$\left(C_{3}\right) 9$ divides this number.
$D:\left(D_{1}\right)$ This number is less than 20;
$\left(D_{2}\right)$ This number is a prime number;
$\left(D_{3}\right)$ 7 divides this number.

This number is - . $\qquad$
Let this number be $n$.
We organize the conditions as follows:
$A:\left(A_{1}\right) n<1000$, i.e., $n>83 \frac{1}{3} ;\left(B_{2}\right) 10 \mid n ;\left(B_{3}\right) n>100$.
$C:\left(C_{1}\right) 4\left|n ;\left(C_{2}\right) 11 n<1000, n<90 \frac{10}{11} ;\left(C_{3}\right) 9\right| n$.
$D:\left(D_{1}\right) n<20 ;\left(D_{2}\right) n$ is a prime number; $\left(D_{3}\right) 7 \mid n$.
First, we notice that $\left(A_{2}\right)$ and $\left(D_{3}\right)$ are completely opposite;
(1) If ( $D_{3} $) is true, $\left(A_{2}\right)$ is false; i.e., $7 \mid n$;
$\left(A_{1}\right)$ and $\left(A_{3}\right)$ must have at least one true statement,

but it cannot be that $\left(A_{1}\right)$ is true and $\left(A_{3}\right)$ is false;
Thus, $\left(A_{3}\right)$ is true; then $n<14$;
Therefore, $\left(D_{1}\right)$ is true;

Thus, ( $D_{2} $) is false;
Notice that the only positive integer that satisfies $n<14$ and $7 \mid n$ is $n=7$, and 7 is a prime number, which contradicts the statement that “( $D_{2}$) $n$ is a prime number” is false.
(2) Therefore, if ( $\left.D_{3}\right)$ is false, $\left(A_{2}\right)$ is true; i.e., $7 \nmid n$;
$\left(A_{1}\right)$ and $\left(A_{3}\right)$ must have at least one false statement,
but it cannot be that $\left(A_{1}\right)$ is true and $\left(A_{3}\right)$ is false;

Thus, ( $A_{1}$ ) is false; then $n \geqslant 12$;

Now, we discuss $\left(D_{1}\right)$:
(1) If ( $D_{1}$ ) is true, i.e., $12 \leqslant n<20$;

Thus, $\left(B_{1}\right)$ and $\left(B_{3}\right)$ are both false;

Then $\left(B_{2}\right)$ is true; then $10 \mid n$.
But there is no positive integer $n$ that satisfies both $12 \leqslant n<20$ and $10 \mid n$.
(2) Therefore, $\left(D_{1}\right)$ is false, i.e., $n \geqslant 20$;

Then $\left(D_{2}\right)$ is true, i.e., $n$ is a prime number;

Thus, $\left(B_{2}\right)$, $\left(C_{1}\right)$, and $\left(C_{3}\right)$ are all false;

Then $\left(C_{2}\right)$ is true, i.e., $20 \leqslant n \leqslant 90$;
Thus, $\left(B_{3}\right)$ is false;

Then $\left(B_{1}\right)$ is true, i.e., $84 \leqslant n \leqslant 90$;
The only number that satisfies $84 \leqslant n \leqslant 90$ and is a prime number is $n=89$, i.e., this number is 89.