14. (17th Japan Mathematical Olympiad) $n$ is a four-digit number with a non-zero tens digit. If the first two digits and the last two digits of $n$ are considered as two two-digit numbers, find all $n$ that satisfy the condition that the product of these two two-digit numbers is a divisor of $n$.


14. Let $A, B$ be the two-digit numbers corresponding to the first two digits and the last two digits of $n$, respectively. Then the problem is transformed into finding all $(A, B)$ such that $A B \mid(100 A+B)$.
Since $A \mid(100 A+B)$, then $A \mid B$.
Let $k=\frac{B}{A}$, because $A, B$ are both two-digit numbers, then $10 \leqslant A<\frac{100}{k}$, so, $k<10$.
Also, $A B\left|(100 A+B) \Rightarrow k A^{2}\right|(100 A+k A) \Rightarrow k A|(100+k) \Rightarrow k|(100+k) \Rightarrow k \mid 100$,
thus $k=1,2,4,5$.
Furthermore, by $A \left\lvert\, \frac{100+k}{k}\right.$ and $10 \leqslant A<\frac{100}{k}$, we get
$(k, A)=(2,17)$ or $(4,13)$.
Therefore, $n=1734$ or 1352.