11. For a $2n$-digit number, if the sum of the first $n$ digits and the last $n$ digits, each considered as an $n$-digit number, squared equals the original $2n$-digit number, then this $2n$-digit number is called a Kabulek strange number. For example, $(30+25)^{2}=3025$, so 3025 is a Kabulek strange number. What are the Kabulek strange numbers among the four-digit numbers?
【Answer】2025, 3025, 9801
【Analysis】Let the first and last two digits of the four-digit number that meets the condition be $a, b$, and $a+b<100$;
Then we have $100 a+b=(a+b)^{2} \Rightarrow(a+b)^{2}-(a+b)=99 a \Rightarrow(a+b)(a+b-1)=99 a$ Since $(a+b)$ and $(a+b-1)$ are coprime, and the product of the two numbers is a multiple of 99, then:
(1) One of the numbers is a multiple of 99, when $a+b=99$ then $(a+b)^{2}=99^{2}=9801$ which meets the requirement;
(2) The two numbers are multiples of 11 and 9 respectively, the conditions that meet are:
$(a+b)(a+b-1)=45 \times 44 \Rightarrow(a+b)^{2}=45^{2}=2025$ which meets the requirement;
$(a+b)(a+b-1)=55 \times 54 \Rightarrow(a+b)^{2}=55^{2}=3025$ which meets the requirement;
In summary, the Kaprekar numbers among the four-digit numbers are: $2025, 3025, 9801$