Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$
Let $M$ be the largest number in $S$.
We categorize numbers $\left\{ 1, 2, \ldots , M-1 \right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups, such that the $i$th group contains two numbers $i$ and $M-i$.
Recall that $M \in S$ and the sum of two numbers in $S$ cannot be equal to $M$, and the sum of numbers in each group above is equal to $S$. Thus, each of the above $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups can have at most one number in $S$.
Therefore,
\begin{align*} |S| & \leq 1 + \left\lfloor \frac{M-1}{2} \right\rfloor \\ & \leq 1 + \left\lfloor \frac{25}{2} \right\rfloor \\ & = 13. \end{align*}
Next, we construct an instance of $S$ with $|S| = 13$.
Let $S = \left\{ 13, 14, \ldots , 25 \right\}$.
Thus, this set is feasible.
Therefore, the most number of elements in $S$ is
$\boxed{\textbf{(B) }13}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)