Let $n=\overline{a b c}$, and assume without loss of generality that $a \geq b \geq c$. We have $k \mid 100 a+10 b+c$ and $k \mid 100 a+10 c+b$, so $k \mid 9(b-c)$. Analogously, $k \mid 9(a-c)$ and $k \mid 9(a-b)$. Note that if $9 \mid n$, then 9 also divides any permutation of $n$ s digits, so $9 \mid f(n)$ as well; ergo, $f(n) \geq 9$, implying that $k \geq 9$. If $k$ is not a multiple of 3 , then we have $k \mid c-a \Longrightarrow k \leq c-a<9$, contradiction, so $3 \mid k$. Let $x=\min (a-b, b-c, a-c)$. If $x=1$, then we have $k \mid 9$, implying $k=9$ - irrelevant to our investigation. So we can assume $x \geq 2$. Note also that $x \leq 4$, as $2 x \leq(a-b)+(b-c)=a-c \leq 9-1$, and if $x=4$ we have $n=951 \Longrightarrow f(n)=3$. If $x=3$, then since $3|k| 100 a+10 b+c \Longrightarrow 3 \mid a+b+c$, we have $a \equiv b \equiv c(\bmod 3)$ (e.g. if $b-c=3$, then $b \equiv c(\bmod 3)$, so $a \equiv b \equiv c(\bmod 3)$ - the other cases are analogous). This gives us the possibilites $n=147,258,369$, which give $f(n)=3,3,9$ respectively. Hence we can conclude that $x=2$; therefore $k \mid 18$. We know also that $k \geq 9$, so either $k=9$ or $k=18$. If $k=18$, then all the digits of $n$ must be even, and $n$ must be a multiple of 9 ; it is clear that these are sufficient criteria. As $n$ 's digits are all even, the sum of them is also even, and hence their sum is 18. Since $a \geq b \geq c$, we have $a+b+c=18 \leq 3 a \Longrightarrow a \geq 6$, but if $a=6$ then $a=b=c=6$, contradicting the problem statement. Thus $a=8$, and this gives us the solutions $n=882,864$ along with their permutations. It remains to calculate the sum of the permutations of these solutions. In the $n=882$ case, each digit is either 8,8 , or 2 (one time each), and in the $n=864$ case, each digit is either 8,6 , or 4 (twice each). Hence the desired sum is $111(8+8+2)+111(8 \cdot 2+6 \cdot 2+4 \cdot 2)=111(54)=5994$.