China Southeastern Mathematical Olympiad

Set $T_0=\{n+1,n+2,\dots ,3n\}$, where $|T_0|=2n$ and the sum of any three elements in $T_0$ is larger than $3n$, that is to say, the sum does not belong to $T_0$. Thus, $\max |T|\ge 2n$.

On the other hand, construct a sequence of sets $$A_0=\{n,2n,3n\},A_k=\{k,2n-k,2n+k\},k=1,2,\dots ,n-1,$$ then $S={\bigcup }_{k=0}^{n-1}{A}_k$. For any subset $T^{\prime }$ in $S$ which has $2n+1$ elements, it must contain a certain $A_k$. If $A_0\subset T^{\prime }$, it contains the element $3n=n+n+n$. If a certain $A_k\subset T^{\prime }$, $k\in \{1,2,\dots ,n-1\}$, it will contain the element $$2n+k=k+k+(2n-k),$$ then $\max |T|<2n+1$. That is, $\max |T|=2n$.