14. Given sets $A, B$ are both sets composed 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 it must be that $\{a, b\}=\{m, n\}$. Define $A+B=\{a+b \mid a \in A, b \in B\}$, try to determine the minimum value of $|A+B|$.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{20}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{16}\right\}$. For each $b_{j} \in B$, define $C_{j}=\left\{a_{i}+b_{j} \mid a_{i} \in A, i=1,2, \cdots, 20\right\}, j=1,2, \cdots, 16$, then we have $A+B=\bigcup_{j=1}^{16} C_{j}$.
We will prove that for any $m \neq n, \left|C_{m} \bigcap C_{n}\right| \leqslant 1$.
Otherwise, there exist $a_{i_{1}} \neq a_{i_{2}}, a_{k_{1}} \neq a_{k_{2}} \in A$ such that $a_{i_{1}}+b_{m}=a_{k_{1}}+b_{n}, a_{i_{2}}+b_{m}=a_{k_{2}}+b_{n}$,
thus $a_{i_{1}}-a_{k_{1}}=a_{i_{2}}-a_{k_{2}} \Rightarrow a_{i_{1}}+a_{k_{2}}=a_{i_{2}}+a_{k_{1}} \Rightarrow \left\{a_{i_{1}}, a_{k_{2}}\right\}=\left\{a_{i_{2}}, a_{k_{1}}\right\}$ $\Rightarrow a_{i_{1}}=a_{k_{1}}, a_{i_{2}}=a_{k_{2}} \Rightarrow b_{m}=b_{n}$, which is a contradiction.
Therefore, by the principle of inclusion-exclusion, we have
$$
|A+B|=\left|\bigcup_{j=1}^{16} C_{j}\right| \geqslant \sum_{j=1}^{16}\left|C_{j}\right|-\sum_{1 \leqslant m<n \leqslant 16}\left|C_{m} \bigcap C_{n}\right|=320-120=200 \text {. }
$$

Take $A=\left\{2,2^{2}, 2^{3}, \cdots, 2^{20}\right\}, B=\left\{2,2^{2}, 2^{3}, \cdots, 2^{16}\right\}$. It is easy to verify that when $m, n, r$ are distinct, $C_{m} \bigcap C_{n}=\left\{2^{m}+2^{n}\right\}, C_{m} \bigcap C_{n} \bigcap C_{r}=\varnothing \Rightarrow|A+B|=200$ satisfies the condition.
In conclusion, the minimum value of $|A+B|$ is 200.