For each $n$ let $f(n)$ be the maximal value of $m$. Readily $f(3) = f(4) = 1$ and $f(5) = 2$. If $n = 6$ each number belongs to at most 2 subsets. Thus, $f(6) \le \frac{6 \cdot 2}{3} = 4$. Therefore $f(6) = 4$ since the example $\{1, 2, 3\}, \{1, 4, 5\}, \{2, 4, 6\}, \{3, 5, 6\}$ works.

Let $n \ge 7$. If some element belongs to all subsets then readily $f(n) \le \lfloor \frac{n-1}{2} \rfloor$. The example $\{1, 2, 3\}, \{1, 4, 5\}, \{1, 6, 7\}, \dots, \{1, n-1, n\}$ for odd $n$ shows that in this case $f(n) = \lfloor \frac{n-1}{2} \rfloor$.

Suppose that no element belongs to all subsets. If some number $a$ belongs to at least 4 subsets, say $S_1, S_2, S_3, S_4$ and some $S_5$ does not contain $a$ then $S_5$ should contain at least 4 elements since it contains at least one element from each $S_i - \{a\}$ for $i = 1, 2, 3, 4$, contradiction.

Suppose that each $a$ belongs to at most 3 subsets. Let us fix one of the subsets: $\{a_1, a_2, a_3\}$. Then there are at most 2 subsets containing each $a_i$, $i = 1, 2, 3$ and there are at most $1+3 \cdot 2 = 7$ subsets. The example for 7 subsets: $\{1, 2, 4\}, \{2, 3, 5\}, \{3, 4, 6\}, \{4, 5, 7\}, \{5, 6, 1\}, \{6, 7, 2\}, \{7, 1, 3\}$.

Thus, $f(n) = \max\{7, \lfloor \frac{n-1}{2} \rfloor\}$.