Let $A$ be a set with $2n$ elements, and let $A_1, A_2...,A_m$ be subsets of $A$e ach one with n elements. Find the greatest possible m, such that it is possible to select these $m$ subsets in such a way that the intersection of any 3 of them has at most one element.
1. **Define the problem and notation:**
   Let \( A \) be a set with \( 2n \) elements, and let \( A_1, A_2, \ldots, A_m \) be subsets of \( A \), each with \( n \) elements. We need to find the greatest possible \( m \) such that the intersection of any three of these subsets has at most one element.

2. **Initial observations:**
   Suppose the elements of \( A \) are \( 1, 2, \ldots, 2n \). If two elements \( i \neq j \) both belong to \( A_1 \) and \( A_2 \), they cannot both belong to any other subset.

3. **Counting elements in subsets:**
   The \( m \) subsets together have \( mn \) elements, possibly with repetitions. Let \( M_i \) denote the number of times element \( i \) appears in the subsets. Then:
   \[
   M_1 + M_2 + \cdots + M_{2n} = mn
   \]

4. **Counting intersections:**
   The number of ways element \( i \) appears in exactly three subsets is given by:
   \[
   \binom{M_i}{3} = \frac{M_i(M_i-1)(M_i-2)}{6}
   \]
   The total number of ways to choose three subsets from \( m \) subsets is:
   \[
   \binom{m}{3} = \frac{m(m-1)(m-2)}{6}
   \]
   Therefore, we must have:
   \[
   \sum_{i=1}^{2n} \binom{M_i}{3} \leq \binom{m}{3}
   \]
   Simplifying, we get:
   \[
   \sum_{i=1}^{2n} \frac{M_i(M_i-1)(M_i-2)}{6} \leq \frac{m(m-1)(m-2)}{6}
   \]
   \[
   \sum_{i=1}^{2n} M_i(M_i-1)(M_i-2) \leq m(m-1)(m-2)
   \]

5. **Minimizing the left-hand side:**
   The left-hand side is minimized when the \( M_i \) differ by at most 1. This is because if \( x > y + 1 \), then:
   \[
   x(x-1)(x-2) + y(y-1)(y-2) > (x-1)(x-2)(x-3) + (y+1)y(y-1)
   \]
   This implies that the \( M_i \) should be as equal as possible.

6. **Case analysis:**
   - **Case 1: \( m = 2a \) (even):**
     \[
     mn = 2an \implies M_i = a \text{ for all } i
     \]
     \[
     2na(a-1)(a-2) \leq 2a(2a-1)(2a-2)
     \]
     Simplifying:
     \[
     na(a-1)(a-2) \leq 2a(2a-1)(2a-2)
     \]
     \[
     n(a-1)(a-2) \leq 4a-2
     \]
     \[
     a(n-4) \leq 2n-2
     \]
     \[
     a \leq \frac{2n-2}{n-4}
     \]
     \[
     m \leq 4 \frac{n-1}{n-4}
     \]

   - **Case 2: \( m = 2a + 1 \) (odd):**
     \[
     mn = (2a+1)n \implies M_1, \ldots, M_n = a \text{ and } M_{n+1}, \ldots, M_{2n} = a+1
     \]
     \[
     n(a(a-1)(a-2) + (a+1)a(a-1)) \leq (2a+1)2a(2a-1)
     \]
     Simplifying:
     \[
     na(a-1)(2a-1) \leq (2a+1)2a(2a-1)
     \]
     \[
     n(a-1) \leq 2(2a+1)
     \]
     \[
     a \leq \frac{n+2}{n-4}
     \]
     \[
     m \leq \frac{3n-2}{n-4}
     \]

7. **Conclusion:**
   For \( n > 16 \), the limit is \( 4 \), which is achievable by taking subsets such as \( (1, 2, \ldots, n), (n+1, \ldots, 2n), (1, 3, \ldots, 2n-1), (2, 4, \ldots, 2n) \).

The final answer is \( \boxed{ m \leq 4 \frac{n-1}{n-4} } \) for even \( m \) and \( m \leq \frac{3n-2}{n-4} \) for odd \( m \).