$n\geq2$ and $E=\left \{ 1,2,...,n \right \}. A_1,A_2,...,A_k$ are subsets of $E$, such that for all $1\leq{i}<{j}\leq{k}$ Exactly one of $A_i\cap{A_j},A_i'\cap{A_j},A_i\cap{A_j'},A_i'\cap{A_j'}$ is empty set. What is the maximum possible $k$?
1. **Define the problem and initial observations:**
   - We are given a set \( E = \{1, 2, \ldots, n\} \) with \( n \geq 2 \).
   - We have subsets \( A_1, A_2, \ldots, A_k \) of \( E \) such that for all \( 1 \leq i < j \leq k \), exactly one of the sets \( A_i \cap A_j \), \( A_i' \cap A_j \), \( A_i \cap A_j' \), \( A_i' \cap A_j' \) is empty.
   - We need to find the maximum possible \( k \).

2. **Symmetry and initial constraints:**
   - By symmetry, if \( \{A_1, \ldots, A_k\} \) is a good family, replacing any \( A_i \) with \( A_i' \) (the complement of \( A_i \) with respect to \( E \)) will still result in a good family.
   - A good family cannot contain two identical sets, sets which are complements of each other, \( \emptyset \), or \( E \) (unless it contains just that subset and no others).

3. **Reduction to sets not containing \( n \):**
   - For every set in a family satisfying the conditions, we can replace it with its complement if it contains \( n \), so no set contains \( n \).
   - This implies \( A_i' \cap A_j' \) will never be \( \emptyset \) for any \( i < j \). Thus, one of the other intersections must be the null set.

4. **Containment and disjointness:**
   - For \( A_i \) and \( A_j \), either one strictly contains the other or they are disjoint.
   - Sort the sets by increasing cardinality, so \( |A_i| \leq |A_j| \) if \( i \leq j \).

5. **Subfamilies and their properties:**
   - Consider subfamilies \( F_m = \{A_1, A_2, \ldots, A_m\} \) (with \( F_0 = \emptyset \)).
   - Let \( S_m = A_1 \cup A_2 \cup \cdots \cup A_m \); let \( B_m \) be the family of subsets \( A_i \) that are not contained in any other subset in \( F_m \) (thus, all the subsets in \( B_m \) are disjoint).
   - Define \( x_m = 2|S_m| - |B_m| \).

6. **Claim: \( x_m < x_{m+1} \) for \( 1 \leq m < k \):**
   - **Case 1: \( S_m = S_{m+1} \):**
     - Then, \( A_{m+1} \subset S_m \).
     - Since for each \( A_i \in B_m \), it is either disjoint with or completely contained in \( A_{m+1} \), we know that \( A_{m+1} \) must be the union of some subsets in \( B_m \).
     - At least two subsets must be involved, or \( A_{m+1} \) would just be identical to a previous subset.
     - Thus, \( B_{m+1} \) will contain \( A_{m+1} \) but not the subsets it's a union of, and \( |B_{m+1}| < |B_m| \), so \( x_m < x_{m+1} \).

   - **Case 2: \( |S_m| < |S_{m+1}| \):**
     - Then \( B_{m+1} \) can only contain one subset not in \( B_m \), namely \( A_{m+1} \).
     - Thus, \( |B_{m+1}| \leq |B_m| + 1 \), so \( x_{m+1} \geq 2(|S_m| + 1) - (|B_m| + 1) = 1 + x_m \), so \( x_{m+1} > x_m \).

7. **Conclusion from the claim:**
   - Thus, the claim is proved. \( x_0 = 0 \) and \( x_k = 2|S_k| - |B_k| \leq 2(n-1) - 1 = 2n-3 \), so \( k \leq 2n-3 \).

8. **Constructing a family that reaches the bound:**
   - Let \( A_i = \{i\} \) for \( 1 \leq i \leq n-1 \) and \( A_{n-2+i} = \{1, 2, \ldots, i\} \) for \( 2 \leq i \leq n-1 \).
   - This construction satisfies the conditions, thus the maximum possible \( k \) is \( 2n-3 \).

The final answer is \( \boxed{2n-3} \)