Three, (25 points) 15 children each have 15 chess pieces. They play a game of "Rock, Paper, Scissors," where each pair of children only has one match, and the loser gives one chess piece to the winner. After the game, the 15 children are divided into two groups, Group A and Group B. The total number of chess pieces in Group A is 63 more than the total number of chess pieces in Group B. Find the maximum and minimum number of chess pieces that the child with the most chess pieces in Group B can have.
Three, let Group A have $x$ people. Then Group B has $15-x$ people.
Let Group A win $y$ times and Group B win $z$ times when the two groups compete. Then $z=x(15-x)-y$.
According to the problem,
$$
\begin{array}{l}
15 x+y-[x(15-x)-y] \\
=15(15-x)-y+[x(15-x)-y]+63 .
\end{array}
$$

Simplifying, we get $y=-\frac{1}{2} x^{2}+72$.
Clearly, $x$ is an even number.
From $y, z \geqslant 0$, we get
$$
\left\{\begin{array}{l}
-\frac{1}{2} x^{2}+72 \geqslant 0, \\
x(15-x)-\left(-\frac{1}{2} x^{2}+72\right) \geqslant 0 .
\end{array}\right.
$$

Solving, we get $6 \leqslant x \leqslant 12$.
Thus, the possible values for $x$ are $6,8,10,12$.
$$
\begin{aligned}
\text { Then }(x, y, z)= & (6,54,0),(8,40,16), \\
& (10,22,28),(12,0,36) .
\end{aligned}
$$

It is easy to see that the total number of chess pieces in Group B is 81.
Therefore, when $x=6$, the minimum value sought is 9;
When $x \neq 6$, the maximum value sought is
$$
15+(15-1)=29 \text {. }
$$

Thus, the maximum number of chess pieces a child in Group B can have is 29, and the minimum is 9.