Three. (50 points) Let $m, n$ be two positive integers (allowing $m$ to equal $n$), and $A, B$ be two sets composed of several real numbers, with $|A|=m, |B|=n$ (allowing $A=B$). The set $A$ satisfies:

If $x, y, u, v \in A$, and $x+y=u+v$, then either $x=y$ and $u=v$, or $\{x, y\}=\{u, v\} (x \neq y$, and $u \neq v)$. Define a set $A+B=\{a+b \mid a \in A, b \in B\}$. Try to find the minimum possible value of $|A+B|$ ($|X|$ denotes the number of elements in set $\boldsymbol{X}$).
Let $A=\left\{a_{1}<a_{2}<\cdots<a_{m}\right\}$, $B=\left\{b_{1}<b_{2}<\cdots<b_{n}\right\}$. The table is listed as follows (see Table 1).

Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
$A$ \\
$B$
\end{tabular} & $a_{1}$ & $a_{2}$ & $\cdots$ & $a_{m-1}$ & $a_{m}$ \\
\hline$b_{1}$ & $a_{1}+b_{1}$ & $a_{2}+b_{1}$ & $\cdots$ & $a_{m-1}+b_{1}$ & $a_{m}+b_{1}$ \\
\hline$b_{2}$ & $a_{1}+b_{2}$ & $a_{2}+b_{2}$ & $\cdots$ & $a_{m-1}+b_{2}$ & $a_{m}+b_{2}$ \\
\hline$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
\hline$b_{n-1}$ & $a_{1}+b_{n-1}$ & $a_{2}+b_{n-1}$ & $\cdots$ & $a_{m-1}+b_{n-1}$ & $a_{m}+b_{n-1}$ \\
\hline$b_{n}$ & $a_{1}+b_{n}$ & $a_{2}+b_{n}$ & $\cdots$ & $a_{m-1}+b_{n}$ & $a_{m}+b_{n}$ \\
\hline
\end{tabular}

In Table 1, the number at the intersection of $a_{i}$ and $b_{j}$ is $a_{i}+b_{j}$, forming a total of $m n$ numbers. Now, we need to delete equal numbers from these $m n$ numbers so that the remaining numbers are all distinct.

Clearly, the $m$ numbers in each row are all distinct, and the $n$ numbers in each column are also all distinct.

Lemma: In Table 1, among any two rows (a total of $2 m$ numbers), it is impossible to have two pairs of numbers that are respectively equal.
Proof of the Lemma: By contradiction.
Consider the $i$-th and $j$-th rows ($i<j$), assume $b_{i}+a_{k_{1}}=b_{j}+a_{l_{1}}$ and $b_{i}+a_{k_{2}}=b_{j}+a_{l_{2}}$ ($k_{1} \neq k_{2}$ and $l_{1} \neq l_{2}$), then $a_{k_{1}}-a_{k_{2}}=a_{l_{1}}-a_{l_{2}}$, i.e., $a_{k_{1}}+a_{l_{2}}=a_{l_{1}}+a_{k_{2}}$.

By the property of $A$, either $a_{k_{1}}=a_{l_{2}}$ and $a_{l_{1}}=a_{k_{2}}$ or $\left\{a_{k_{1}}, a_{l_{2}}\right\}=\left\{a_{l_{1}}, a_{k_{2}}\right\}$ ($a_{k_{1}} \neq a_{l_{2}}$ and $a_{l_{1}} \neq a_{k_{2}}$).

From the former, we get $a_{k_{1}}=a_{l_{1}}=a_{k_{2}}=a_{l_{2}}$, thus $k_{1}=k_{2}$, which contradicts the assumption $k_{1} \neq k_{2}$.

From the latter, we get $a_{k_{1}}=a_{l_{1}}$ and $a_{l_{2}}=a_{k_{2}}$ (since $k_{1} \neq k_{2}$, so $a_{k_{1}} \neq a_{k_{2}}$). Thus, $b_{i}-b_{j}=a_{l_{1}}-a_{k_{1}}=0$, i.e., $i=j$, which contradicts the assumption $i<j$.
Returning to the original problem.
By the lemma, at most one number can be deleted from any two rows (deleting one of the two equal numbers), so at most $\mathrm{C}_{n}^{2}$ numbers can be deleted from Table 1, leaving at least $m n-\mathrm{C}_{n}^{2}$ numbers that are all distinct, i.e., $|A+B| \geqslant m n-C_{n}^{2}$.
(1) When $m \geqslant n$, take $a_{i}=3^{i}$ ($i=1,2, \cdots, m$) and $b_{j}=3^{j}$ ($j=1,2, \cdots, n$), $A$ has the property given in the problem, in this case $|A+B|=m n-\mathrm{C}_{n}^{2}$, so the minimum value of $|A+B|$ is $m n-\mathrm{C}_{n}^{2}$ (when $m \geqslant n$).
(2) When $m<n$, consider Table 2.
Table 2
Notice that the $m$ rows of $b_{1}, b_{2}, \cdots, b_{m}$ and the $m$ columns of $a_{1}, a_{2}, \cdots, a_{m}$ form an $m \times m$ square (marked with a black frame), the remaining part is a $(n-m) \times m$ rectangle, the numbers in the $m$-th column of this rectangle are
$\Delta_{m+1}=a_{m}+b_{m+1}, \cdots, \Delta_{n}=a_{m}+b_{n}$.
Let $B_{1}=\left\{b_{1}, b_{2}, \cdots, b_{m}\right\}, B_{2}=\left\{b_{m+1}, \cdots, b_{n}\right\}$, then by the conclusion of (1) we have
$$
\begin{array}{l}
|A+B|=|\{a+b \mid a \in A, b \in B\}| \\
=\left|\left\{a+b \mid a \in A, b \in B_{1}\right\} \cup\left\{a+b \mid a \in A, b \in B_{2}\right\}\right| \\
\geqslant\left|\left\{a+b \mid a \in A, b \in B_{1}\right\} \cup\left\{a_{m}+b \mid b \in B_{2}\right\}\right| \\
=1\left\{a+b \mid a \in A, b \in B_{1}\right\}\left|+1\left\{\Delta_{m+1}, \cdots, \Delta_{n}\right\}\right| \\
\left.\geqslant m^{2}-C_{m}^{2}+(n-m)=n+C_{m}^{2} \text { (when } m<n \text { ) }\right. .
\end{array}
$$

Furthermore, it can be shown by example that the equality in the above inequality can be achieved. Therefore, the minimum value of $|A+B|$ is $n+C_{m}^{2}$ (when $m<n$).
In summary, we have
$$
|A+B|_{\min }=\left\{\begin{array}{l}
m n-C_{n}^{2}=n\left(m-\frac{n-1}{2}\right), m \geqslant n ; \\
n+C_{m}^{2}=n+\frac{m(m-1)}{2}, m<n .
\end{array}\right.
$$

Note: When $m=n$, $m n-\mathrm{C}_{n}^{2}=n+\mathrm{C}_{m}^{2}$.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510006)