One. (20 points) Let $m, n$ be positive integers, and $m \neq 2$. The quadratic function $y=x^{2}+(3-m t) x-3 m t$ intersects the $x$-axis at two points with a distance of $d_{1}$ between them. The quadratic function $y=-\dot{x}^{2}+(2 t-n) x+2 n t$ intersects the $x$-axis at two points with a distance of $d_{2}$ between them. If $d_{1} \geqslant d_{2}$ holds for all real numbers $t$, find the values of $m$ and $n$.
$$
\begin{array}{l}
\text { Because the roots of the quadratic equation } \\
x^{2}+(3-m t) x-3 m t=0 \\
\text { are } m t \text { and } -3, \text { so } d_{1}=|m t+3|; \\
\text { and the roots of the quadratic equation } \\
-x^{2}+(2 t-n) x+2 n t=0 \\
\text { are } 2 t \text { and } -n, \text { so } d_{2}=|2 t+n|. \\
\text { Hence } d_{1} \geqslant d_{2} \Leftrightarrow|m t+3| \geqslant|2 t+n| \\
\Leftrightarrow(m t+3)^{2} \geqslant(2 t+n)^{2} \\
\Leftrightarrow\left(m^{2}-4\right) t^{2}+(6 m-4 n) t+9-n^{2} \geqslant 0 .
\end{array}
$$

Given that $m^{2}-4 \neq 0$, and inequality (1) holds for all real numbers $t$, we have:
$$
\begin{array}{l}
\left\{\begin{array}{l}
m^{2}-4>0, \\
\Delta=(6 m-4 n)^{2}-4\left(m^{2}-4\right)\left(9-n^{2}\right) \leqslant 0
\end{array}\right. \\
\Rightarrow\left\{\begin{array} { l } 
{ m > 2 , } \\
{ 4 ( m n - 6 ) ^ { 2 } \leqslant 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m>2, \\
m n=6 .
\end{array}\right.\right. \\
\text { Therefore }\left\{\begin{array} { l } 
{ m = 3 , } \\
{ n = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
m=6, \\
n=1
\end{array}\right.\right.
\end{array}
$$