imc

Let $M$ be the largest number in $S$. We categorize numbers $\left\{1,2,\dots ,M-1\right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\lfloor \frac{M-1}{2}\rfloor$ 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 $\lfloor \frac{M-1}{2}\rfloor$ groups can have at most one number in $S$. Therefore, $$\begin{array}{rl}|S|& \le 1+\lfloor \frac{M-1}{2}\rfloor \\ & \le 1+\lfloor \frac{25}{2}\rfloor \\ & =13.\end{array}$$ Next, we construct an instance of $S$ with $|S|=13$. Let $S=\left\{13,14,\dots ,25\right\}$. Thus, this set is feasible. Therefore, the most number of elements in $S$ is $\boxed{13}$.