[Solution] Let $x$ represent the number of bricks in the first pile, and $y$ represent the number of bricks in the second pile. Let $z$ be the number of bricks transferred from the second pile to the first pile according to the problem's conditions. At this point, the following equations hold:
$$\begin{array}{l} 2(x-100)=y+100, \\ x+z=6(y-z) \end{array}$$

From (1), we get $y=2 x-300$.
Substituting into (2), we get $11 x-7 z=1800$.
That is, $4 x+7(x-z)=1800$.
Therefore, $x-z$ is a multiple of 4. Let $x-z=4 t, t \in \mathbb{Z}$, substituting into (3) we get
$$x+7 t=450, x=-7 t+450$$

Thus,
$$\begin{array}{l} y=2 x-300=-14 t+600 \\ z=x-4 t=-11 t+450 \end{array}$$

The numbers $x, y, z$ are all natural numbers, so they should satisfy the inequalities:
$$\left\{\begin{array}{l} -7 t+450>0 \\ -14 t+600>0 \\ -11 t+450>0 \end{array}\right.$$

Solving this system of inequalities, we get $t \leqslant 40$.
The value of $x$ decreases as $t$ increases, so the minimum value of $x$ is obtained when $t$ is at its maximum allowable value, which is $t=40$. At this point, $x=170, y=40, z=10$; checking the values of $x, y, z$ obtained, we find that they satisfy all the conditions of the problem. Therefore, the minimum number of bricks in the first pile is 170, and at this time, the number of bricks in the second pile is 40.