Four, $n^{2}(n \geqslant 4)$ positive numbers are arranged in $n$ rows and $n$ columns,
\begin{tabular}{llllll}
$a_{11}$ & $a_{12}$ & $a_{13}$ & $a_{14}$ & $\cdots$ & $a_{1 n}$ \\
$a_{21}$ & $a_{22}$ & $a_{23}$ & $a_{24}$ & $\cdots$ & $a_{2 n}$ \\
$a_{31}$ & $a_{32}$ & $a_{33}$ & $a_{34}$ & $\cdots$ & $a_{3 n}$ \\
$a_{41}$ & $a_{42}$ & $a_{43}$ & $a_{41}$ & $\cdots$ & $a_{4 n}$ \\
$\cdots \cdots$ & & & & & \\
$a_{n 1}$ & $a_{n 2}$ & $a_{n 3}$ & $a_{n 4}$ & $\cdots$ & $a_{n n}$
\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
$$
a_{11}+a_{22}+a_{33}+a_{44}+\cdots+a_{n n}
$$
Let the common difference of the first row be $d$ and the common ratio of each column be $q$. Then the common difference of the fourth row is $d q^{3}$. Thus, we have the system of equations:
$$
\left\{\begin{array}{l}
a_{24}=\left(a_{11}+3 d\right) q=1, \\
a_{42}=\left(a_{11}+d\right) q^{3}=\frac{1}{8}, \\
a_{43}=\frac{1}{8}+d q^{3}=\frac{3}{16},
\end{array}\right.
$$

Solving this system, we get
$$
a_{11}=d=q= \pm \frac{1}{2} .
$$

Since all $n^{2}$ numbers are positive, we have
$$
a_{11}=d=q=\frac{1}{2} .
$$

For any $1 \leqslant k \leqslant n$, we have
$$
a_{k k}=a_{1 k} q^{k-1}=\left[a_{11}+(k-1) d\right] q^{k-1}=k \frac{1}{2^{k}},
$$

Thus,
$$
S=\frac{1}{2}+2 \frac{1}{2^{2}}+\cdots+n \frac{1}{2^{n}},
$$

And since
$$
\frac{1}{2} S=\frac{1}{2^{2}}+2 \frac{1}{2^{3}}+\cdots+n \frac{1}{2^{n+1}},
$$

Subtracting these, we get
$$
\frac{1}{2} S=\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}-n \frac{1}{2^{n+1}}=\frac{\frac{1}{2}\left(1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}}-n \frac{1}{2^{n+1}} .
$$

Therefore,
$$
S=2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}} .
$$