11. Consider a $2n$-digit number, where the first $n$ digits and the last $n$ digits are each treated as an $n$-digit number. If the square of the sum of these two $n$-digit numbers is exactly equal to the $2n$-digit number, then this $2n$-digit number is called a Kabulek (Kabulek) strange number. For example, $(30+25)^{2}=3025$, so 3025 is a Kabulek strange number. Please find which four-digit numbers are Kabulek strange numbers?
【Analysis】Let the first two digits of the number be $x$, and the last two digits be $y$. Thus, we have $(x+y)^{2}=100 x+y=x+y+99 x$, which means: $(x+y)$
$$
(x+y-1)=99 x \text {, }
$$

From this, we can see that one of $x+y$ and $x+y-1$ is a multiple of 9, and the other is a multiple of 11 (of course, depending on the number of digits, it could also be other factors), thus finding the three numbers that satisfy the condition: 45, 55, and 99, and then calculating their squares.

【Solution】Solution: Let the first two digits of the number be $x$, and the last two digits be $y$. Thus, we have $(x+y)^{2}=100 x+y=x+y+99 x$, which means: $(x+y)(x+y-1)=99 x$,
From this, we can see that one of $x+y$ and $x+y-1$ is a multiple of 9, and the other is a multiple of 11 (of course, depending on the number of digits, it could also be other factors),

We can find the three numbers that satisfy the condition: 45, 55, and 99, and their squares are $2025, 3025, 9801$;
That is: $(20+25)^{2}=2025$;
$(30+25)^{2}=3025$;
$(98+1)^{2}=9801$;
Answer: The Kaprekar numbers in four-digit numbers are: $2025, 3025, 9801$.