14. Given 10 positive integers, the sum of any 9 of them takes only the following 9 different values: $86, 87, 88, 89$, $90, 91, 93, 94, 95$, then among these 10 positive integers, when arranged in descending order, find the sum of the 3rd and the 7th numbers.
14. Let the 10 positive integers, arranged in descending order, be $x_{1}, x_{2}, \cdots, x_{10}$, and let $S=\sum_{i=1}^{10} x_{i}$. Then, the sum of any 9 of these numbers, $x_{1}, x_{2}, \cdots, x_{10}$, has only the following $C_{10}=10$ possibilities: $S_{i}=S-x_{i}(i=1,2, \cdots, 10)$. Since $S_{1}, \cdots, S_{10}$ take only 9 different values, there must be two that are equal. Let these two numbers be the same value $k$, then
$$
\begin{array}{l}
\sum_{i=1}^{10} S_{i}=10 S-\sum_{i=1}^{10} x_{i}=9 S=86+87+88+89+90+91+93+94+95+k=810+3+k . \\
\therefore a \mid 3+k, \text { and } 86 \leqslant k \leqslant 95 . \\
\therefore 3+k=90, \text { so } k=87, \therefore S=100 . \\
x_{1}=S-S_{1}=100-86=14, x_{2}=x_{3}=13, x_{4}=12, x_{5}=11, x_{6}=10, x_{7}=9, x_{8}=7, x_{9}=6, x_{10}=5 . \\
\therefore x_{3}+x_{7}=22 .
\end{array}
$$