14. $n^{2}(n \geqslant 4)$ positive numbers are arranged in $n$ rows and $n$ columns,
\begin{tabular}{llllll}
$a_{11}$ & $a_{12}$ & $a_{13}$ & $a_{14}$ & $\cdots \cdots$ & $a_{1 n}$ \\
$a_{21}$ & $a_{22}$ & $a_{23}$ & $a_{24}$ & $\cdots \cdots$ & $a_{2 n}$ \\
$a_{31}$ & $a_{32}$ & $a_{33}$ & $a_{34}$ & $\cdots \cdots$ & $a_{3 n}$ \\
$a_{41}$ & $a_{42}$ & $a_{43}$ & $a_{44}$ & $\cdots \cdots$ & $a_{4 n}$ \\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots \cdots$ & $\cdots$ \\
$a_{n 1}$ & $a_{n 2}$ & $a_{n 3}$ & $a_{n 4}$ & $\cdots \cdots$ & $a_{n n}$
\end{tabular}

where the numbers in each row form an arithmetic sequence, and the numbers in each column form a geometric sequence, with all common ratios being equal. Given $a_{24}=1, a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$, find $a_{11}+a_{22}+\cdots+a_{n n}$.
Let the common difference of the $n$-th row be $d_{n}$, and the common ratio of each column be $q$.
Then $d_{4}=a_{43}-a_{42}=\frac{1}{16} \Rightarrow a_{4 n}=a_{43}+\frac{1}{16}(n-3)=\frac{n}{16}$.
Also, $a_{44}=\frac{1}{4} \Rightarrow q^{2}=\frac{a_{44}}{a_{24}}=\frac{1}{4} \Rightarrow q=\frac{1}{2}$, thus $a_{n n}=a_{4 n} \cdot \frac{1}{2^{n-4}}=\frac{n}{2^{n}}$.
Let $\frac{n}{2^{n}}=\frac{k n+b}{2^{n-1}}-\frac{k(n+1)+b}{2^{n}}=\frac{k n+b-k}{2^{n}} \Rightarrow k=b=1$,
i.e., $\frac{n}{2^{n}}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^{n}} \Rightarrow \frac{1}{2^{1}}=2-\frac{3}{2^{1}}, \frac{2}{2^{2}}=\frac{3}{2^{1}}-\frac{4}{2^{2}}, \cdots, \frac{n}{2^{n}}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^{n}}$.
Therefore, $a_{11}+a_{22}+\cdots+a_{n n}=2-\frac{n+2}{2^{n}}\left(n \in \mathbf{N}^{*}\right)$.