19. (Mathematics for Middle School, Issue 4, 2004, Olympiad Problem 137) Try to find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two different elements $a$ and $b$ in $S$, there exists another element $c$ in $S$ such that the greatest common divisor of $a+b$ and $c$ is 1;
(3) For any two different elements $a$ and $b$ in $S$, there exists another element $c$ in $S$ such that the greatest common divisor of $a+b$ and $c$ is greater than 1.
19. Construct 50 arrays:
$(1,100),(2,99), \cdots,(50,51)$, the sum of the two numbers in each array is 101.
Since 101 is a prime number, there does not exist an element $c$ in $S$ such that the greatest common divisor of 101 and $c$ is greater than 1. Therefore, it is impossible for $S$ to contain both numbers from the same array. By the pigeonhole principle, the number of elements in set $S$ is no more than 50.
On the other hand, we construct the set $A=\{2,1,3,5,7, \cdots, 95,97\}$. This set contains 2 and 49 odd numbers less than 98.
We now show that set $A$ satisfies the given conditions.
For any two elements $a$ and $b$ in set $A$:
(i) If $a=2$, then $b$ is an odd number.
If $b=1$, it is easy to see that there exists an element $c$ in $A$ that satisfies the given conditions.
If $3 \leqslant b \leqslant 95$, then the greatest common divisor of the element 1 in $A$ and $a+b$ is 1, and the greatest common divisor of the element $b+2$ in $A$ and $a+b$ is $b+2$, which is greater than 1.
If $b=97$, it is easy to see that there exists an element $c$ in $A$ that satisfies the given conditions.
(ii) If $a$ and $b$ are not equal to 2, then $a$ and $b$ are both odd numbers, and $a+b$ is even. Thus, the greatest common divisor of $a+b$ and 2 is 2, which is greater than 1, and $a+b$ must be coprime with one of the numbers 1, 89, or 91.
Therefore, set $A$ satisfies the given conditions.
Thus, the maximum number of elements in set $S$ is 50.