Let $ m,n$ be positive integers, and let $ F$ be a family of $ m$-element subsets of $ \{1,2,...,n\}$ satisfying $ A\cap B \not \equal{} \emptyset$ for all $ A,B\in F$. Determine the maximum possible number of elements in $ F$.
To determine the maximum possible number of elements in \( F \), we need to consider two cases based on the relationship between \( m \) and \( n \).

### Case 1: \( 2m > n \)
In this case, any two \( m \)-element subsets of \( \{1, 2, \ldots, n\} \) must intersect. This is because the total number of elements in the union of any two such subsets is at most \( 2m \), which is greater than \( n \). Therefore, any two subsets must share at least one element.

1. **Counting all possible \( m \)-element subsets:**
   \[
   \binom{n}{m}
   \]
   Since all \( m \)-element subsets intersect, the family \( F \) can contain all possible \( m \)-element subsets.

### Case 2: \( 2m \leq n \)
In this case, it is possible to have two \( m \)-element subsets that do not intersect. Therefore, we need to find a way to construct \( F \) such that all subsets in \( F \) intersect.

1. **Constructing a family \( F \) with \( \binom{n-1}{m-1} \) subsets:**
   - Fix an element, say \( 1 \), and consider all \( m \)-element subsets that include \( 1 \).
   - The remaining \( m-1 \) elements can be chosen from the remaining \( n-1 \) elements.
   \[
   \binom{n-1}{m-1}
   \]
   This construction ensures that all subsets in \( F \) intersect because they all share the element \( 1 \).

2. **Proving that \( \binom{n-1}{m-1} \) is the maximum:**
   - Consider the total number of \( m \)-element subsets, which is \( \binom{n}{m} \).
   - Note that \( \binom{n-1}{m-1} = \frac{m}{n} \binom{n}{m} \).
   - If \( n = 2m \), each \( m \)-element subset can be paired with a unique disjoint \( m \)-element subset. Thus, \( F \) can contain at most half of all possible subsets.
   - For general \( n \), partition the set of all \( m \)-element subsets into groups where each group contains \( l \) disjoint subsets (where \( l = \left\lfloor \frac{n}{m} \right\rfloor \)). By symmetry, each subset is equally represented, and at most one subset from each group can be in \( F \).

3. **Handling non-multiples of \( m \):**
   - Construct a sub-family containing \( n \) subsets such that any valid \( F \) can contain at most \( m \) of them.
   - Example: \(\{1, 2, \ldots, m\}, \{2, 3, \ldots, m+1\}, \ldots, \{n, 1, \ldots, m-1\}\).
   - Renaming elements and using symmetry arguments, we conclude that at most \( \frac{m}{n} \) of all possible subsets can be in \( F \).

Thus, the maximum possible number of elements in \( F \) is:
- \( \binom{n}{m} \) if \( 2m > n \)
- \( \binom{n-1}{m-1} \) if \( 2m \leq n \)

The final answer is \( \boxed{\binom{n}{m} \text{ for } 2m > n \text{ and } \binom{n-1}{m-1} \text{ for } 2m \leq n} \)