Four. (50 points) Let $S \subseteq\{1,2,3, \cdots, 100\}$. It is known that for any two distinct elements $a, b$ in $S$, there exist a positive integer $k$ and two distinct elements $c, d$ in $S$ (which may equal $a$ or $b$), such that $c<d$, and $a+b=c^{k} d$. Find the maximum number of elements in the set $S$.

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Four. (50 points) Let $S \subseteq\{1,2,3, \cdots, 100\}$. It is known that for any two distinct elements $a, b$ in $S$, there exist a positive integer $k$ and two distinct elements $c, d$ in $S$ (which may equal $a$ or $b$), such that $c<d$, and $a+b=c^{k} d$. Find the maximum number of elements in the set $S$.
Four, Solution: Take $S_{0}=\{1,2,3,4,5,7,9,11,13, \cdots, 89,91\}$ as a 48-element set. Below, we verify that the set $S_{0}$ satisfies the problem's conditions:
If we take two numbers $a, b \in S_{0}$ with the same parity, then $a+b$ is even, and $4 \leqslant a+b \leqslant 89+91$.
When $a+b \neq 2^{t}(t=2,3, \cdots, 7)$, take $c=2, k$ as the positive integer that exactly divides $a+b$ by $2^{k}$, at this time $\frac{a+b}{2^{k}}$ must be an odd number less than 91 and not equal to 1, let it be $d$.
When $a+b=2^{t}(t=3,4, \cdots, 7)$, we can take $k=t-2, c=2, d=4$.
When $a+b=4$, we can take $k=c=1, d=4$.
If we take two numbers $a, b \in S_{0}$ with different parity, then $a+b$ is odd, and $3 \leqslant a+b \leqslant 4+91=95$.
When $a+b \in\{3,4, \cdots, 91\}$, we can take $k=c=1, d=a+b$.
When $a+b=93$, we can take $k=1, c=3, d=31$.
When $a+b=95$, we can take $k=1, c=5, d=19$.
Thus, $S_{0}$ satisfies the problem's conditions.
Below, we prove that the number of elements in a set $S$ that satisfies the problem's conditions does not exceed 48.
Let $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, and $a_{1}1+\left(t-2+i_{0}\right)=1+a_{i_{0}}$, which is a contradiction! Therefore, $m=1$. Thus, $1+a_{i_{0}}=a_{l}$, at this time $a_{i_{0}}+1=a_{l}=t-2+\left(i_{0}+1\right)$,
So when $i=i_{0}+1$, it holds. By induction, Lemma 2 is true.
Lemma 3: If a prime $p \notin S$, then there do not exist $m, l, 1 \leqslant m<l \leqslant n$, such that $a_{m}+a_{l}=p$.
Proof: When a prime $p \notin S$, it is clear that $p$ cannot be expressed in the form $c^{k} d$, where $c, d \in S$. Combining the known conditions, there do not exist such $a_{m}, a_{l}$.
Below, we prove $n \leqslant 48$.
By Lemma 1, we know $a_{1}=1$.
When $a_{2}=2$, by Lemma 2, we know $a_{3}=3, a_{4}=4$.
Using Lemma 3 for the prime 101, we know $97,98,99,100 \notin S$. Using Lemma 3 for the prime 97, we know: among the 48 pairs of numbers $(1,96),(2,95), \cdots$, $(48,49)$, at most one number from each pair belongs to $S$, hence $n \leqslant 48$.

When $3 \leqslant a_{2} \leqslant 6$, using Lemma 2, we know $6,7,8,9 \notin S$. Since $97+6=103,98+9=99+8=100+7=107$, using Lemma 3 for the primes 103, 107, we know $97,98,99,100 \notin S$. Similar to the case when $a_{2}=2$, hence $n \leqslant 48$.

When $a_{2} \geqslant 7$, using Lemma 3 for the prime 107, we know: among the 47 pairs of numbers $(7,100),(8,99), \cdots,(53,54)$, at most one number from each pair belongs to $S$, hence $n \leqslant 47 \pm 1=48$, in summary, the maximum number of elements in $S$ is 48.