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

where the numbers in each row form an arithmetic sequence, and the numbers in each column form a geometric sequence, and all the common ratios are equal. Given $a_{24}=1$,
$$
a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16} \text {, find } a_{11}+a_{22}+a_{33}+a_{44}+\cdots+a_{n n} \text {. }
$$
$$
2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}}
$$

Four, 【Analysis and Solution】Let the common difference of the first row sequence be $d$, and the common ratio of each column sequence be $q$, then the common difference of the fourth row sequence is $d q^{3}$, thus we can obtain 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 of equations, we get
$$
a_{11}=d=q= \pm \frac{1}{2} \text {, }
$$

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
$$
a_{k k}=a_{1 k} q^{k-1}=\left[a_{11}+(k-1) d\right] q^{k-1}=k \frac{1}{2^{k}} \text {, }
$$

Thus, $S=\frac{1}{2}+2 \frac{1}{2^{2}}+\cdots+n \frac{1}{2^{n}}$.
Also, $\frac{1}{2} S=\frac{1}{2^{2}}+2 \frac{1}{2^{3}}+\cdots+n \frac{1}{2^{n+1}}$,
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, $S=2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}}$.