7. If the set $S=\{1,2, \cdots, n\}$ has three-element subsets $A_{1}, A_{2}, \cdots, A_{m}(m>1)$ satisfying:
(1) Each two-element subset of $S$ is contained in exactly one $A_{i}$ $(1 \leqslant i \leqslant m)$;
(2) $A_{i}$ and $A_{j}(i \neq j, 1 \leqslant i, j \leqslant m)$ have exactly one common element.
Then $m=$ . $\qquad$
7. 7 .

Each $A_{i}$ contains 3 binary subsets, and $S$ has a total of $\mathrm{C}_{n}^{2}$ binary subsets. Therefore,
$$
3 m=\mathrm{C}_{n}^{2} \text {. }
$$

For an element $k$ in $S$, there are $n-1$ binary subsets in $S$ that contain $k$.

By condition (1), each of these is contained in exactly one $A_{i}$, and each $A_{i}$ containing $k$ includes two binary subsets that contain $k$, so there are $\frac{n-1}{2}$ $A_{i}$'s containing $k$.

Considering $A_{i}$ as points, and connecting each pair of points with an edge, a graph is formed with $\mathrm{C}_{m}^{2}$ edges.

By condition (2), the edge between $A_{i}$ and $A_{j}$ can be labeled with the unique common element $k$ of $A_{i}$ and $A_{j}$, and the edge labeled $k$ appears exactly $\mathrm{C}_{\frac{n-1}{2}}^{2}$ times, thus,
$$
\mathrm{C}_{m}^{2}=n \mathrm{C}_{\frac{n-1}{2}}^{2} .
$$

Solving equations (1) and (2) yields $(m, n)=(1,3),(7,7)$.
$$
\text { Since } m>1 \text {, we have } m=7 \text {. }
$$