[Proof] Let the 1953-digit number starting from a certain position be
$$A=\overline{a_{1} a_{2} \cdots a_{n-1} a_{n}} \quad(n=1953),$$

Then, by the given condition, $27 \mid A$.
We only need to prove that $B=\overline{a_{n} a_{1} a_{2} \cdots a_{n-1}}$ is divisible by 27, and then, each time we move one digit from the end to the beginning, we can get all numbers divisible by 27.
Consider
$$10 B+a_{n}$$
$$\begin{aligned} 10 B+a_{n} & =a_{n} \cdot 10^{n}+A \\ & =a_{n} \cdot \underbrace{99 \cdots 9}_{1953 \uparrow}+A+a_{n} . \end{aligned}$$

Since $99 \cdots 9$ and $A$ are both divisible by 27, then
1953
$$\begin{array}{l} 10 B+a_{n} \equiv a_{n} \quad(\bmod 27), \\ 10 B \equiv 0 \quad(\bmod 27) . \\ \end{array}$$

Also, $(10,27)=1$, so $B \equiv 0 \quad(\bmod 27)$.
Thus, $B$ is divisible by 27.