4. (Mathematical Olympiad Problems, Intermediate Mathematics, Issue 6, 2002) Find all integers $k$ such that the quadratic equation $k x^{2}-2(3 k-1) x+9 k-1=0$ has at least one integer root.

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4. From the original equation, we get $k(x-3)^{2}=1-2 x$. Since $x$ is an integer root, it is clear that $x \neq 3, k \neq 0$. Therefore, $k=\frac{1-2 x}{(x-3)^{2}}$. Given that $k$ is an integer, we have $|1-2 x| \geqslant(x-3)^{2}$.
If $x \leqslant 0$, then $1-2 x \geqslant x^{2}-6 x+9$, which simplifies to $(x-2)^{2}+4 \leqslant 0$. This is impossible.
If $x>0$, then $2 x-1 \geqslant x^{2}-6 x+9$, which simplifies to $(x-4)^{2} \leqslant 6$, or $|x-4| \leqslant \sqrt{6}$. Thus, $|x-4| \leqslant 2$, so $x$ can only be $2, 4, 5, 6$.

Upon individual verification, we find that only $x=2$ (corresponding to $k=-3$) and $x=4$ (corresponding to $k=-7$) satisfy the conditions. Therefore, $k=-3$ or $-7$.