3. (43rd American High School Mathematics Examination) Let $S$ be a subset of the set $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements of $S$ cannot be divisible by 7. What is the maximum number of elements that $S$ can have?
For two different natural numbers $a$ and $b$, if $7 \times (a+b)$, then the sum of their remainders when divided by 7 is not 0. Therefore, the set $\{1,2, \cdots, 50\}$ can be divided into 7 subsets based on the remainders when divided by 7, where each element in $A_{i}$ has a remainder of $i$ when divided by 7 ($i=1,2, \cdots, 6$), then
\[
\begin{aligned}
A_{0} & =\{7,14,21,28,35,42,49\}, \\
A_{1} & =\{1,8,15,22,29,36,43,50\}, \\
A_{2} & =\{2,9,16,23,30,37,44\}, \\
A_{3} & =\{3,10,17,24,31,38,45\}, \\
A_{1} & =\{4,11,18,25,32,39,46\}, \\
A_{5} & =\{5,12,19,26,33,40,47\}, \\
A_{6} & =\{6,13,20,27,34,41,48\}.
\end{aligned}
\]

According to the problem:
(1) $S$ can contain at most one element from $A_{0}$;
(2) If $S$ contains one element from $A_{i}$, then it can contain all elements from this set, but it cannot contain any elements from $A_{7-i}$;
(3) $A_{1}$ contains 8 elements, while the other subsets contain only 7 elements, so the largest subset $S$ must contain all elements from $A_{1}$. In summary, the largest subset $S$ has $1+8+7+7=23$ elements.