Three. (25 points) In a population of 1000 individuals numbered $0,1,2, \cdots, 999$, they are sequentially divided into 10 groups: Group 0 includes numbers $0,1, \cdots, 99$, Group 1 includes numbers $100,101, \cdots, 199$, and so on. Now, one number is to be selected from each group to form a sample of size 10, with the following rule: If a number $x$ is arbitrarily chosen from Group 0, then the numbers in the subsequent groups are obtained by shifting positions, i.e., the last two digits of the number drawn from Group $k$ are the last two digits of $x + a k$.
(1) When $x=24, a=33$, write down the 10 numbers in the sample;
(2) If $a=33$ and one of the 10 numbers in the sample has the last two digits 87, find the sum of all possible values of $x$ that satisfy the condition;
(3) Does there exist a two-digit positive integer $a$ such that at least two of the 10 numbers in the sample have the same last two digits? If it exists, find the value of $a$; if not, explain the reason.
(1) When $x=24, a=33$, the values of $x+a k$ corresponding to $k$ taking the values $0,1, \cdots, 9$ are
$$
24,57,90,123,156,189,222,255,288,321 \text {. }
$$

Therefore, the 10 numbers of the sample drawn are
$$
24,157,290,323,456,589,622,755,888,921 \text {. }
$$
(2) Since a number with the last two digits 87 can appear in each group, and when $k$ takes the values $0,1, \cdots, 9$, the values of $33 k$ are
$$
0,33,66,99,132,165,198,231,264,297 \text {, }
$$

Thus, all possible values of $x$ are
$$
87,54,21,88,55,22,89,56,23,90 \text {. }
$$

Their sum is 585.
(3) Suppose the last two digits of the numbers drawn from the $i$-th group and the $j$-th group are the same, i.e., the last two digits of $x+a i$ and $x+a j$ are the same. Therefore, the last two digits of $a|i-j|$ must be 0, and $1 \leqslant|i-j| \leqslant 9$.

Since $a$ is a two-digit positive integer, all values of $a$ that satisfy the condition are $20,25,40,50,60,75,80$.