15. (6 points) There are 4 different digits that can form a total of 18 different four-digit numbers, which are arranged in ascending order. The first number is a perfect square, and the second-to-last four-digit number is also a perfect square. What is the sum of these two numbers? $\qquad$
【Solution】Solution: Let the four digits be $03$; the smallest (first and second are: $c b 0 a$,

Below, we start the discussion from the value of $c$ (in combination with $0<a<b<c$):
Based on the characteristics of the units digit of square numbers: $c=4,5,6,9$, when $c=4$: only $32 \times 32=1024$; but 4201 is not a square number, so it is excluded. When $c=5$: $45 \times 45=2025 ; 55 \times 55=3025$ do not meet the criteria, so they are excluded. When $c=6$: all are excluded as they do not meet the criteria. When $c=9$: $33 \times 33=1089 ; 9801=99 \times 99$ meet the conditions; the smallest is 1089, the second to last is 9801, and thus we can find the sum of these two numbers. The sum of these two numbers is: $1089+9801=10890$. Therefore, the answer is: 10890.