11. (10 points) Find all four-digit numbers $n$ that satisfy the following conditions:
(1) The first and third digits of $n$ are the same;
(2) The second and fourth digits of $n$ are the same;
(3) The product of the digits of $n$ is a divisor of $n^2$.
【Analysis】First, set the form of the unknowns, then discuss the relationship between the two numbers, while paying attention to the fact that this four-digit number is a multiple of 101 and also a prime number. Find the numerical relationship and enumerate accordingly.

【Solution】Solution: According to the problem, we have:
Let $n=\overline{\mathrm{ab} \mathrm{ab}}=101 \overline{\mathrm{ab}}$, and $a^{2} b^{2}$ is a divisor of $n^{2}$, so $a b$ is a divisor of $n$. Since 101 is a prime number, i.e., $a b$ is a divisor of $101 \overline{\mathrm{ab}}$, and because $a b$ is coprime with 101, $a b$ is a divisor of $\overline{\mathrm{ab}}$. Therefore, $a b \mid(10 a+b)$, which implies $a \mid b$ and $b \mid 10 a$, so $b=a, 2 a$ or $5 a$,

Thus, the value of $\overline{\mathrm{ab}}$ can be $11, 12, 15, 24, 36$.
Answer: The four-digit numbers are: $1111, 1212, 1515, 2424, 3636$.