Solution:
We will prove that the maximum value that $T_{A}$ can attain is $4$. Let $A=\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\}$ be a set of five positive integers such that $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. Call a triple $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ good if $x_{i}+x_{j}+x_{k}$ divides $S_{A}$. None of the triples $(3,4,5),(2,4,5),(1,4,5),(2,3,5),(1,3,5)$ is good, since, for example
$$x_{5}+x_{3}+x_{1}\mid S_{A} \Rightarrow x_{5}+x_{3}+x_{1}\mid x_{2}+x_{4}$$
which is impossible since $x_{5}>x_{4}$ and $x_{3}>x_{2}$. Analogously we can show that any triple of form $(x, y, 5)$ where $y>2$ isn't good.

By above, the number of good triples can be at most $5$ and only triples $(1,2,5),(2,3,4)$, $(1,3,4),(1,2,4),(1,2,3)$ can be good. But if triples $(1,2,5)$ and $(2,3,4)$ are simultaneously good we have that:
$$x_{1}+x_{2}+x_{5} \mid x_{3}+x_{4} \Rightarrow x_{5}<x_{3}+x_{4}$$
and
$$x_{2}+x_{3}+x_{4} \mid x_{1}+x_{5} \Rightarrow x_{2}+x_{3}+x_{4} \leqslant x_{1}+x_{5} < x_{1}+x_{3}+x_{4}<x_{2}+x_{3}+x_{4},$$
which is impossible. Therefore, $T_{A} \leqslant 4$.

Alternatively, one can prove the statement above by adding up the two inequalities $x_{1}+x_{2}+x_{4}<x_{3}+x_{4}$ and $x_{2}+x_{3}+x_{4}<x_{1}+x_{5}$ that are derived from the divisibilities.

To show that $T_{A}=4$ is possible, consider the numbers $1,2,3,4,494$. This works because $6|498,7|497,8|496$, and $9|495$.