4・209 There are two small piles of bricks. If 100 bricks are taken from the first pile and placed in the second pile, then the second pile will be twice as large as the first pile. If a certain number of bricks are taken from the second pile and placed in the first pile, then the first pile will be six times the size of the second pile. How many bricks are there in the first pile at a minimum? And determine the number of bricks in the second pile at this time.
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. The following equations hold:
$$
\begin{array}{l}
2(x-100)=y+100 \\
x+z=6(y-z)
\end{array}
$$

From (1), we get $y=2x-300$.
Substituting into (2), we get $11x-7z=1800$.
This can be rewritten as $4x+7(x-z)=1800$.
Thus, $x-z$ is a multiple of 4. Let $x-z=4t, t \in \mathbb{Z}$, substituting into (3) gives
$$
x+7t=450, \quad x=-7t+450 \text{, }
$$

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

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

Solving this system of inequalities, we get $t \leqslant 40$.
The value of $x$ decreases as $t$ increases, so the minimum value of $x$ occurs when $t$ is at its maximum allowed value, which is $t=40$. Thus, $x=170, y=40, z=10$; checking these values, 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.