5. $n+1$.

Solution: Let $T$ contain a total of $k$ elements. Subtracting the other elements from the largest element in $T$, the differences obtained are not in $T$. Therefore, we have
$k+(k-1) \leqslant 2 n+1$, which means $k \leqslant n+1$. Additionally, the subset $\{n+1, n+1, \cdots, 2 n+1\}$ of $S$ contains $n+1$ elements, which clearly meets the requirements of the problem. Hence, the maximum value of $k$ is $n+1$.