11. (15 points) There is a strange four-digit number (the first digit is not 0), which is a perfect square, and the sum of its digits is also a perfect square. When this four-digit number is divided by the sum of its digits, the result is still a perfect square, and the number of its divisors is exactly equal to the sum of its digits, which is also a perfect square. If the digits of this four-digit number are all different, what is this four-digit number?


【Answer】Solution:
The number of divisors of a square number must be odd, and the digit sum of a square number and 1 is obviously impossible.
The digit sum can only be $1$, $9$, or $2$,
A square number with 25 divisors must end in 25 (which means this number contains the prime factor 5), and must be in the form of $a^{4} \times b^{4}$ or $c^{4}$ (obviously too large, abandon), at least $2^{4} \times 5^{4}=10000$, not a four-digit number,
So the digit sum of this square number is 9 (which means this number contains the prime factor 3), and has 9 divisors, then it must be in the form of $a^{2} \times b^{2}$ or $c^{8}$,
$$
3^{2} \times 17^{2}=2601
$$

Answer: This four-digit number is 2601.