Let $n$ be an integer greater than 1. For a positive integer $m$, let $S_{m} = \{ 1, 2, \ldots, mn \}$. Suppose that there exists a $2n$-element set $T$ such that:
(a) each element of $T$ is an $m$-element subset of $S_{m}$;
(b) each pair of elements of $T$ shares at most one common element; and
(c) each element of $S_{m}$ is contained in exactly two elements of $T$.

We aim to determine the maximum possible value of $m$ in terms of $n$.

First, we show that $m \leq 2n - 1$. By condition (b), there are at most $\binom{2n}{n}$ elements of $S_{m}$ which are in 2 sets in $T$. However, by condition (c), every element of $S_{m}$ is in 2 sets in $T$, so $|S_{m}| = mn \leq \binom{2n}{n} = \frac{(2n)(2n-1)}{2} = n(2n-1)$, and thus $m \leq 2n - 1$.

Now, to see that $2n - 1$ is achievable, consider a complete graph $K_{2n}$. Note that this $K_{2n}$ has $n(2n-1)$ edges, so we label the edges with distinct elements of $S_{2n-1}$. Let $T$ be the family of sets formed by taking each vertex and creating the set of labels of incident edges. Any two sets in $T$ have exactly one element in common—the label of the edge between them—so condition (b) is satisfied. Each element of $S_{m}$ is in exactly two elements of $T$: the two endpoints of the edge with the given label, satisfying condition (c). Condition (a) is also satisfied because every vertex in a $K_{2n}$ has degree $2n-1 = m$, so every element of $T$ has $m$ elements. Finally, since a $K_{2n}$ has $2n$ vertices, $|T| = 2n$, as desired.

Thus, the maximum possible value of $m$ is:
$$\boxed{2n - 1}$$