Let's determine all integer values of $m$ and $n$ for which the roots of the following equation are also integers:

$$
(2 m-3)(n-1) x^{2}+(2 m-3)(n-1)(m-n-4) x-2(2 m-3)(n-1)(m-n-2)-1=0
$$
Since $m$ is an integer, $2m-3 \neq 0$. $n-1$ also cannot be 0, otherwise from the equation we would conclude $-1=0$, which is impossible. Let's add 1 to both sides of the equation and divide both sides by $(2m-3)(n-1) \neq 0$.

$$
x^{2}+(m-n-4) x-2(m-n-2)=\frac{1}{(2m-3)(n-1)}
$$

If $x_{1}$ and $x_{2}$ are the roots of the above equation, then

$$
\begin{gathered}
x_{1}+x_{2}=-(m-n-4) \quad \text { and } \\
x_{1} x_{2}=-\left[2(m-n-2)+\frac{1}{(2m-3)(n-1)}\right]
\end{gathered}
$$

Since $m$ and $n$ are integers, $x_{1}+x_{2}$ is also an integer. If the product of the roots is an integer, then $\frac{1}{(2m-3)(n-1)}$ must also be an integer. This is true if the denominator is +1 or -1. Let's examine the two cases:

(A) If the denominator is 1, then $2m-3=1$ and $n-1=1$, or $2m-3=-1$ and $n-1=-1$.

In the first case, $m=2$ and $n=2$, in the second case, $m=1$ and $n=0$.

Let's check the roots of the equation for the obtained values of $m$ and $n$.

If $m=2$ and $n=2$, then the equation (1) is $x^{2}-4 x+3=0$, from which $x_{1}=3, x_{2}=1$. The roots are integers in this case.

If $n=1$ and $n=0$, then the equation (1) is $x^{2}-3 x+1=0$ and $x_{1,2}=\frac{3 \pm \sqrt{5}}{2}$, which are not integers.

(B) If the denominator is -1, then $2m-3=1$ and $n-1=-1$, or $2m-3=-1$ and $n-1=1$. In the first case, $m=2, n=0$, in the second case, $m=1$ and $n=2$.

In the case of $m=2$ and $n=0$, the roots of the obtained equation will be integers $\left(x^{2}-2 x+1=0 ; x_{1,2}=1\right)$, but in the second case, the equation has no real roots.

Therefore, the equation (1) will have integer roots only in the cases $m=2, n=2$ and $m=2, n=0$.