14. Let $n \geq 3$ be a positive integer, and let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ numbers. Denote the pairwise sums as $b_{i j}=a_{i}+a_{j}(i>j)$, resulting in the following table:
$$
\begin{array}{l}
b_{21} \\
b_{31} \quad b_{32} \\
\text {........................... } \\
b_{n 1} \quad b_{n 2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots b_{n, n-1} \\
\end{array}
$$

If any $k$ numbers are chosen from the above table, they can uniquely determine the $n$ numbers $a_{1}, a_{2}, \cdots, a_{n}$. Find the minimum value of $k$.
【Analysis】: (1) When $n=3$, it is clear that $b_{21}=a_{2}+a_{1}, b_{32}=a_{3}+a_{2}, b_{31}=a_{3}+a_{1}$ can uniquely determine $a_{1}, a_{2}, a_{3}$, at this time $k=3$. (5 points)
(2) When $n=4$, it is clear that $k \geq C_{3}^{2}+1=4$, otherwise the sum of any two out of three numbers cannot determine the fourth number.
When $k=4$, if the values of $b_{21}, b_{31}, b_{42}, b_{43}$ are known, it is also impossible to determine $a_{1}, a_{2}, a_{3}, a_{4}$.
When $k=5$, if any five of the values $b_{21}, b_{31}, b_{32}, b_{41}, b_{42}, b_{43}$ are known. Suppose the value of $b_{21}$ is unknown, then by $b_{31}, b_{32}, b_{42}, b_{43}$, we can determine $a_{3}\left(=\frac{1}{2}\left(b_{32}+b_{43}-b_{42}\right)\right)$, thus uniquely determining $a_{1}, a_{2}, a_{3}, a_{4}, \cdots \cdots$ (10 points)
(3) When $n \geq 5$, it is clear that when $k \geq C_{n-1}^{2}+1$, the minimum value is achieved. Below is the proof.
(a) When $n=5$, $k=C_{4}^{2}+1=7$, i.e., if 7 of the $b_{i j}$ are known, then there must exist an index $s, b_{i s}$ (or $b_{j s}$) that appears at most 2 times and at least 1 time. In fact, 7 $b_{i j}$ have 14 indices, and if each of $1,2,3,4,5$ appears 3 times or more, there would be 15 indices, which is impossible. Therefore, according to (2), by at least 5 of the $b_{i j}, i, j \in\{1,2,3,4,5\} \backslash\{s\}$, we can uniquely determine $a_{i}, i \in\{1,2,3,4,5\} \backslash\{s\}$, and then by the $b_{i s}$ (or $b_{j s}$) that appears at least once, we can uniquely determine $a_{s}$.
(20 points)
(b) When $n \geq 6$, we prove by mathematical induction that $k=C_{n-1}^{2}+1$. When $k$ $b_{i j}$ are taken, there must exist an index $s, b_{i s}$ (or $b_{j s}$) that appears at most $n-2$ times (since $2 k<n(n-1)$), then $b_{i j}, i, j \in\{1,2,3, \cdots, n\} \backslash\{s\}$ has at least
$$
k-(n-2)=C_{n-1}^{2}-n+3 \geq C_{n-1}^{2}+1
$$

By induction, these $b_{i j}$ can uniquely determine $a_{i}, i \in\{1,2,3, \cdots, n\} \backslash\{s\}$, and then by $b_{i s}$ (or $b_{j s}$) determine $a_{s}$.
(30 points)