One, (20 points) In a school's donation activity for the "Hope Project", the total donation amount from $m$ boys and 11 girls in Class A is equal to the total donation amount from 9 boys and $n$ girls in Class B, which is $(m \cdot n + 9 m + 11 n + 145)$ yuan. It is known that each person's donation amount is the same and is an integer number of yuan. Find the donation amount per person.
$$
-\because m n+9 m+11 n+145=(m+11)(n+9)+
$$
46, by the given
$$
\begin{array}{l}
m+11 \mid(m n+9 m+11 n+145), \\
n+9 \mid(m n+9 m+11 n+145)
\end{array}
$$

and
$$
\begin{array}{l}
m+11=n+9, \\
\therefore m+11|46, n+9| 46 .
\end{array}
$$
$\because m, n$ are non-negative integers,
$$
\begin{array}{l}
\therefore m+11 \geqslant 11, n+9 \geqslant 9, \\
\text { and } \because 46=1 \times 46=2 \times 23, \\
\therefore m+11=n+9=46 \text { or } m+11=n+9=23 .
\end{array}
$$

When $m+11=n+9=46$,
$$
\begin{array}{l}
m n+9 m+11 n+145 \\
=46 \times 46+46=47 \times 46 .
\end{array}
$$

At this time, the amount each person carries is
$$
47 \times 46 \div 46=47 \text { ( } \pi \text { ). }
$$

When $m+11=n+9=23$,
$$
\begin{array}{l}
m n+9 m+11 n+145 \\
=23 \times 23+46=25 \times 23 .
\end{array}
$$

At this time, the amount each person carries is
$$
25 \times 23 \div 23=25 \text { (yuan). }
$$

In summary, the amount each person carries is 47 yuan or 25 yuan.