14. (20 points) Given sets $A$ and $B$ are both sets of positive integers, and $|A|=20,|B|=16$. Set $A$ satisfies the following condition: if $a, b, m, n \in A$, and $a+b=m+n$, then $\{a, b\}=\{m, n\}$. Define
$$
A+B=\{a+b \mid a \in A, b \in B\} .
$$

Determine the minimum value of $|A+B|$.
14. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{20}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{16}\right\}, \\
C_{j}=\left\{a_{i}+b_{j} \mid i=1,2, \cdots, 20\right\},
\end{array}
$$

where $j=1,2, \cdots, 16$.
Thus, $A+B=\bigcup_{j=1}^{16} C_{j}$.
We will prove: $\left|C_{m} \cap C_{n}\right| \leqslant 1(m \neq n)$.
In fact, suppose there exist $C_{m} 、 C_{n}(m \neq n),\left|C_{m} \cap C_{n}\right|$
$\geqslant 2$. Then there exist $k_{1} 、 k_{2} 、 l_{1} 、 l_{2}$, such that
$$
\text { and } \begin{array}{l}
a_{k_{1}}+b_{m} 、 a_{k_{2}}+b_{m} 、 a_{l_{1}}+b_{n} 、 a_{l_{2}}+b_{n} \in C_{m} \cap C_{n}, \\
a_{k_{1}}+b_{m}=a_{k_{2}}+b_{n}, a_{l_{1}}+b_{m}=a_{l_{2}}+b_{n}, \\
a_{k_{1}} \neq a_{k_{2}}, a_{l_{1}} \neq a_{l_{2}} .
\end{array}
$$

Then $a_{k_{1}}-a_{k_{2}}=a_{l_{1}}-a_{l_{2}}$
$$
\Rightarrow a_{k_{1}}+a_{l_{2}}=a_{l_{1}}+a_{k_{2}},
$$

which implies $\left\{a_{k_{1}}, a_{l_{1}}\right\}=\left\{a_{k_{2}}, a_{l_{2}}\right\}$.
By conclusion (1), we know $a_{k_{1}}=a_{l_{1}}, a_{k_{2}}=a_{l_{2}}$, which is impossible, otherwise,
$$
a_{k_{1}}+b_{m} \in C_{n}, a_{k_{2}}+b_{n} \in C_{m}
$$