, 2002

Solution:
First consider the ones digit $a$ of $n$; we claim that $a=1$. Certainly $a$ cannot be even, for then $5 n$ would be divisible by $10$. If $a$ is $5$, $7$, or $9$, then $2 n$ has an even ones digit, while its most significant digit is $1$. If $a$ is $3$, then $4 n$ has an even ones digit but most significant digit $1$. Thus $a=1$ is the only possibility. Moreover $9 n$ has the same number of digits as $n$, for otherwise $9 n$ would have most significant digit $1$ but least significant digit $9$, which is forbidden.

Now suppose $n$ has at least one digit that is neither a zero nor a one. Let $b$ be the leftmost (i.e., most significant) such digit, so that the left end of the decimal representation of $n$ looks like
$$\underline{a_{1}} \cdots \underline{a_{r}} \underline{b} \cdots$$
for some $r \geq 1$ and digits $a_{i} \in \{0,1\}$. When $n$ is multiplied by $9$, there will be a carry out of the column containing $b$. In particular, the $r^{\text{th}}$ digit from the left in $9 n$ will not be $9 a_{r}$. But the right end of the decimal representation of $n$ is
$$\cdots \underline{a_{r}} \cdots \underline{a_{1}} ;$$
because each $a_{i}$ is $0$ or $1$, there are no carries out of the first $r-1$ columns, so the $r^{\text{th}}$ digit from the right in $9 n$ will be $9 a_{r}$. Thus $9 n$ is not a palindrome, a contradiction. This completes the proof.