Solution:
Let $r(m)$ denote the number obtained from $m$ by reversing the digits.

We show first that $k$ cannot be divisible by $2$ or $5$. It cannot be divisible by both, for then it ends in a zero and hence $r(k) < k$ and so is not divisible by $k$ (contradiction). So if $5$ divides $k$, then the last digit of $k$ must be $5$. Since $r(k)$ is divisible by $5$ its last digit must also be $5$, so the first digit of $k$ is $5$. But now $3k$ has first digit $1$ ($3 \times 5 > 10$ and $3 \times 6 < 20$), so $r(3k)$ has last digit $1$ and cannot be divisible by $5$. Contradiction. If $2$ divides $k$, then every multiple of $k$ must be even. So the last digit of $r(k)$ must be even and hence the first digit of $k$ must be $2$, $4$, $6$, or $8$. If $2$, then $5k$ has first digit $1$, so $r(2k)$ is odd. Contradiction. Similarly, if the first digit is $4$, $3k$ has first digit $1$; if $6$, then $5k$ has first digit $3$; if $8$, then $2k$ has first digit $1$. Contradiction. So $k$ is not divisible by $2$ or $5$.

Suppose $k = 10^{n} a_{n} + \ldots + a_{0}$. $k$ divides $r(k)$, so $a_{0} \geq 1$. Hence $(10^{n+1} - 1)k = 10^{2n+1} a_{n} + \ldots + 10^{n+1} a_{0} - (10^{n} a_{n} + \ldots + a_{0}) = 10^{2n+1} a_{n} + \ldots + 10^{n+1}(a_{0} - 1) + 10^{n} c_{n} + \ldots + 10 c_{1} + (c_{0} + 1)$, where $c_{i} = 9 - a_{i}$. The reverse of this, $10^{2n+1}(c_{0} + 1) + 10^{2n} c_{1} + \ldots + 10^{n+1} c_{n} + 10^{n}(a_{0} - 1) + \ldots + a_{n}$, is also divisible by $k$. So is the reverse of $k$, $10^{n} a_{0} + \ldots + a_{n}$ and hence also their difference: $10^{n}(10^{n+1}(c_{0} + 1) + 10^{n} c_{1} + \ldots + 10 c_{n} - 1)$. $k$ has no factors $2$ or $5$, so $k$ must divide $10^{n+1}(c_{0} + 1) + 10^{n} c_{1} + \ldots + 10 c_{n} - 1$. Adding $10k$, we find that $k$ also divides $10^{n+2} + 10^{n} 9 + \ldots + 10 \times 9 - 1 = 1099\ldots 989$ ($n-2$ consecutive $9$s$) = 11(10^{n+1} - 1)$.

We can now carry out exactly the same argument starting with $(10^{n+2} - 1)k$. This leads to $k$ dividing $10^{n+2}(c_{0} + 1) + \ldots + 10^{2} c_{0} + 10 \times 9 - 1$ and hence also $10^{n+3} + 10^{n+1} 9 + \ldots + 10^{2} 9 + 10 \times 8 + 9 = 11(10^{n+2} - 1)$. Subtracting $10$ times this from the previous number we conclude that $k$ must divide $11(10^{n+1} - 1) - 11(10^{n+1} - 10) = 99$.

Finally, we note that any factor of $99$ has the required property. For $3$ and $9$ divide a number if and only if they divide its digit sum. So if $m$ is divisible by $3$ or $9$, then the number formed by any rearrangement of its digits is also divisible by $3$ or $9$. $m$ is divisible by $11$ if and only if the difference between the sums of alternate digits is divisible by $11$, so if $m$ is divisible by $11$, then so is its reverse.