Given a positive integer $n(n>1)$, for a positive integer $m$, the set $S_{m}=\{1,2, \cdots, m n\}$. The family of sets $\mathscr{T}$ satisfies the following conditions:
(1) Each set in $\mathscr{T}$ is an $m$-element subset of $S_{m}$;
(2) Any two sets in $\mathscr{T}$ have at most one common element;
(3) Each element of $S_{m}$ appears in exactly two sets in $\mathscr{T}$.
Find the maximum value of $m$.

Try to solve the problem.
Second, the maximum value of $m$ is $2n-1$.
First, estimate the upper bound of $m$.
On one hand, consider the set
$$
\begin{aligned}
U= & \left\{\left(i,\left\{T_{j}, T_{k}\right\}\right) \mid i \in S_{m}, T_{j} \neq T_{k},\right. \\
& \left.T_{j}, T_{k} \in \mathscr{T}, i \in T_{j} \cap T_{k}\right\} .
\end{aligned}
$$

By condition (3), for any element $i$ in $S_{m}$, there is exactly one pair $\left\{T_{j}, T_{k}\right\}$ such that $\left(i,\left\{T_{j}, T_{k}\right\}\right) \in U$. Therefore, $|U| = |S_{m}| = mn$.
On the other hand, consider the set
$$
V=\left\{\left(i, T_{j}\right) \mid i \in S_{m}, T_{j} \in \mathscr{T}, i \in T_{j}\right\} .
$$

By condition (3), $|V| = 2mn$.
By condition (1), $|V| = m|\mathscr{T}|$.
Thus, $|\mathscr{T}| = 2n$.
By condition (2), $\left|T_{j} \cap T_{k}\right| \leqslant 1$.
Therefore, $|U| \leqslant \mathrm{C}_{2n}^{2}$. Hence,
$$
mn \leqslant \mathrm{C}_{2n}^{2} = \frac{2n(2n-1)}{2} = n(2n-1),
$$

which implies $m \leqslant 2n-1$.
Next, use mathematical induction to construct an example where $m = 2n-1$.
For $n=2, m=3, S_{m}=\{1,2, \cdots, 6\}$, the set family
$$
\mathscr{T}=\{\{1,2,3\},\{3,5,6\},\{1,4,5\},\{2,4,6\}\}
$$

satisfies the conditions.
Assume that when $n=k$,
$$
\begin{array}{l}
S_{m}=\{1,2, \cdots, k(2k-1)\}, \\
\mathscr{T}=\left\{T_{1}, T_{2}, \cdots, T_{2n}\right\}
\end{array}
$$

satisfies the conditions.
When $n=k+1$, $m=2k+1$,
$$
S_{m}=\left\{1,2, \cdots, k(2k-1), a_{1}, a_{2}, \cdots, a_{4k+1}\right\},
$$

where $a_{i}=k(2k-1)+i$ for $i=1,2, \cdots, 4k+1$.
$$
\text{Let } \mathscr{F}^{\prime}=\bigcup_{i=1}^{2k}\left\{T_{i} \cup\left\{a_{2i-1}, a_{2i}\right\}\right\} \cup\left\{T_{2k+1}, T_{2k+2}\right\},
$$

where $T_{2k+1}=\left\{a_{1}, a_{3}, \cdots, a_{4k-1}, a_{4k+1}\right\}$,
$$
T_{2k+2}=\left\{a_{2}, a_{4}, \cdots, a_{4k}, a_{4k+1}\right\}.
$$

After verification, the set family $\mathscr{T}^{\prime}$ satisfies the conditions.
Thus, by the principle of induction, $m=2n-1$ can be achieved.
In conclusion, the maximum value of $m$ is $2n-1$.