For what real values of $k$ are the solutions of the equation in $x$
$$
(6-k)(9-k) x^{2}-(117-15 k) x+54=0
$$

all integers?
It is known that when $k=6$ or 9, the solutions of the original equation are all integers.
When $k \neq 6$ and $k \neq 9$, the original equation can be transformed into
$$
[(6-k) x-9][(9-k) x-6]=0 \text {. }
$$

Solving, we get $x_{1}=\frac{9}{6-k}=\frac{3}{2-\frac{k}{3}}, x_{2}=\frac{6}{9-k}=\frac{2}{3-\frac{k}{3}}$.
Since $x_{1}, x_{2}$ are integers, it is easy to see that $\frac{k}{3}$ is a rational number. Let $\frac{k}{3} = \frac{m}{n}$, where $m$ is an integer, $n$ is a positive integer, and $(m, n)=1$. Substituting into $x_{1}, x_{2}$, we get
$$
\begin{array}{c}
x_{1}=\frac{3 n}{2 n-m}, \quad x_{2}=\frac{2 n}{3 n-m} . \\
\begin{aligned}
\because(n, 2 n-m) & =(n, 2 n-(2 n-m)) \\
& =(n, m)=1 ;
\end{aligned} \\
(n, 3 n-m)=(n, 3 n-(3 n-m))=(n, m)=1 . \\
\therefore 2 n-m|3, \quad 3 n-m| 2 .
\end{array}
$$

Thus, $2 n-m= \pm 1, \pm 3 ; 3 n-m= \pm 1, \pm 2$.
Noting that $2 n-m<3 n-m$, we combine the various cases as shown in the table below:
\begin{tabular}{c|c|c|c|c|c|c|c}
\hline $2 n-m$ & 1 & -1 & -1 & -3 & -3 & -3 & -3 \\
\hline $3 n-m$ & 2 & 1 & 2 & -1 & 1 & -2 & 2 \\
\hline$n$ & 1 & 2 & 3 & 2 & 4 & 1 & 5 \\
\hline$m$ & 1 & 5 & 7 & 7 & 11 & 5 & 13 \\
\hline
\end{tabular}

Therefore, $\frac{k}{3}=1, \frac{5}{2}, \frac{7}{3}, \frac{7}{2}, \frac{11}{4}, 5, \frac{13}{5}$.
Then $k=3, \frac{15}{2}, 7, \frac{21}{2}, \frac{33}{4}, 15, \frac{39}{5}$.
In summary, when $k$ is $3, 6, 7, \frac{15}{2}, \frac{39}{5}, \frac{33}{4}, 9, \frac{21}{2}, 15$, the solutions of the original equation are all integers.
(Chen Kuanhong, Xiongshi Middle School, Yueyang County, Hunan Province, 414113)