Find all integers $k$ such that one of the roots of the quadratic equation $k x^{2}-2(3 k-1) x+9 k-1=0$ is a reduced fraction with a denominator of 1999.
Solution: From the problem, we know that $k \neq 0, \Delta=4(1-5 k) \geqslant 0$, so $k<0$, and the two roots are
$$
x_{1,2}=\frac{3 k-1 \pm \sqrt{1-5 k}}{k}=3+\frac{-1 \pm \sqrt{1-5 k}}{k} .
$$

Let $\sqrt{1-5 k}=m$, where $m$ is a positive integer and $m \geqslant 4$, then
$$
k=\frac{1-m^{2}}{5} \text {. }
$$

Thus, $m \equiv 1(\bmod 5)$ or $m \equiv 4(\bmod 5)$. Therefore, the last digit of $m$ can only be $1,4,6,9$.
-From (1) and (2) we get
$$
\left\{\begin{array}{l}
x_{1}=3+\frac{5}{m-1}, \\
x_{2}=3-\frac{5}{m}+1
\end{array}\right.
$$

From (3), we can divide $m$ into four cases for discussion:
(1) When $m=10 a+1$, $x_{1}=3+\frac{1}{2 a}, x_{2}=3-$ $\frac{5}{10 a+2}$, which does not meet the requirements.
(2) When $m=10 a+4$, $x_{1}=3+\frac{5}{10 a+3}, x_{2}=3$ $-\frac{1}{2 a+1}$, we can only set $2 a+1=1999$ in $x_{2}$, so $a=$ 999. Thus, $m=9994$.
Hence $k=\frac{1-9994^{2}}{5}=-9993 \times 1999$.
(3) When $m=10 a+6$, $x_{1}=3+\frac{1}{2 a+1}, x_{2}=3-$ $\frac{5}{10 a+7}$, we can only set $2 a+1=1999$ in $x_{1}$, so $a=$ 999. Thus, $m=9996$. Hence $k=\frac{1-9996^{2}}{5}=-9997 \times$ 1999.
(4) When $m=10 a+9$, $x_{1}=3+\frac{5}{10 a+8}, x_{2}=3$ $-\frac{1}{4 a+2}$, which does not meet the requirements.

In summary, the required $k=-1999 \times 9993$ or $-1999 \times 9997$.