一、(20 points) Find all real numbers $k$ such that the quadratic equation in $x$
$$
k x^{2}-2(3 k-1) x+9 k-1=0
$$

has two integer roots.
Let the equation be
$$
k x^{2}-2(3 k-1) x+9 k-1=0
$$

with two integer roots $x_{1}, x_{2}$, and $x_{1} \leqslant x_{2}$. Then
$$
x_{1}+x_{2}=6-\frac{2}{k}, x_{1} x_{2}=9-\frac{1}{k} \text {. }
$$

Eliminating $\frac{1}{k}$ from the above two equations, we get
$$
2 x_{1} x_{2}-x_{1}-x_{2}=12 \text {, }
$$

which simplifies to $\left(2 x_{1}-1\right)\left(2 x_{2}-1\right)=5^{2}$, and $2 x_{1}-1 \leqslant 2 x_{2}-1$.
$$
\left\{\begin{array} { l } 
{ 2 x _ { 1 } - 1 = - 2 5 , } \\
{ 2 x _ { 2 } - 1 = - 1 ; }
\end{array} \left\{\begin{array}{l}
2 x_{1}-1=-5, \\
2 x_{2}-1=-5 .
\end{array}\right.\right.
$$

Solving these, we get $\left\{\begin{array}{l}x_{1}=1, \\ x_{2}=13 ;\end{array}\left\{\begin{array}{l}x_{1}=3 \\ x_{2}=3 ;\end{array}\left\{\begin{array}{l}x=-12, \\ x_{2}=0 ;\end{array}\left\{\begin{array}{l}x_{1}=-2, \\ x_{2}=-2 .\end{array}\right.\right.\right.\right.$
Upon verification, $x_{1}=x_{2}=3$ is an extraneous root, hence
$$
k=\frac{1}{9-x_{1} x_{2}}=-\frac{1}{4}, \frac{1}{9}, \frac{1}{5} .
$$