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

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

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