$11 \cdot 4$ Given that using a set of weights in groups of 6 can balance 63 weights of consecutive natural numbers, find this set of weights.
(21st All-Soviet Union Mathematical Olympiad, 1987)
[Solution] Let the weights of the 6 weights we are looking for, arranged in non-decreasing order, be $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{6}$. The sum of some of the numbers $x_{1}, x_{2}, \cdots, x_{6}$ can be expressed as
$$
\varepsilon_{1} x_{1}+\varepsilon_{2} x_{2}+\cdots+\varepsilon_{6} x_{6}
$$

where $\varepsilon_{i}=0,1$, and at least one $\varepsilon_{i} \neq 0$.
There are exactly $2^{6}-1=63$ expressions of the form (1).
According to the problem, these 63 numbers can be written as $x_{0}+1, x_{0}+2, \cdots, x_{0}+$ 63. Since all expressions of (1) are distinct, it must be that $x_{1}<x_{2}<\cdots<x_{6}$.
And we have
$$
\begin{array}{l}
x_{0}+1=x_{1}, \quad x_{0}+62=x_{2}+x_{3}+\cdots+x_{6}, \\
x_{0}+63=x_{1}+x_{2}+\cdots+x_{6} .
\end{array}
$$

From this, we get
$$
x_{1}=1 \text {. }
$$

Thus,
$$
x_{0}=0, \quad x_{2}=2, \quad x_{1}+x_{2}=3, \quad x_{3}=4 .
$$

Assume the numbers $x_{1}, x_{2}, \cdots, x_{k}(k \leqslant 5)$ have already been found, so the expression $\varepsilon_{1} x_{1}+\varepsilon_{2} x_{2}+$ $\cdots+\varepsilon_{k} x_{k}\left(\varepsilon_{i}=0,1\right)$ can give the numbers $1,2, \cdots, 2^{k}-1$, then $x_{k+1}=2^{k}$, and the expression $\varepsilon_{1} x_{1}+\varepsilon_{2} x_{2}+\cdots+\varepsilon_{k+1} x_{k+1}$ for the same $\varepsilon_{i}$ represents the numbers $1,2, \cdots, 2^{k+1}-1$.
Thus, we can find sequentially:
$$
\begin{array}{l}
x_{1}=1, \quad x_{2}=2, \quad x_{3}=4, \quad x_{4}=2^{3}=8, \\
x_{5}=2^{4}=16, \quad x_{6}=2^{5}=32 .
\end{array}
$$