Let $P$ be the set of all $2012$ tuples $(x_1, x_2, \dots, x_{2012})$, where $x_i \in \{1,2,\dots 20\}$ for each $1\leq i \leq 2012$. The set $A \subset P$ is said to be decreasing if for each $(x_1,x_2,\dots ,x_{2012} ) \in A$ any $(y_1,y_2,\dots, y_{2012})$ satisfying $y_i \leq x_i (1\leq i \leq 2012)$ also belongs to $A$. The set $B \subset P$ is said to be increasing if for each $(x_1,x_2,\dots ,x_{2012} ) \in B$ any $(y_1,y_2,\dots, y_{2012})$ satisfying $y_i \geq x_i (1\leq i \leq 2012)$ also belongs to $B$. Find the maximum possible value of $f(A,B)= \dfrac {|A\cap B|}{|A|\cdot |B|}$, where $A$ and $B$ are nonempty decreasing and increasing sets ($\mid \cdot \mid$ denotes the number of elements of the set).
1. **Base Case: \( n = 1 \)**

   Consider the set \( P \) of all 1-tuples where each element \( x_i \) can take any value from 1 to 20. Let \( A \) be a decreasing set and \( B \) be an increasing set. Suppose:
   \[
   A = \{1, 2, \ldots, a+c\}
   \]
   and
   \[
   B = \{20-b-c+1, \ldots, 20\}
   \]
   Here, \( |A \cap B| = c \), \( |A| = a+c \), and \( |B| = b+c \). Then,
   \[
   f(A, B) = \frac{|A \cap B|}{|A| \cdot |B|} = \frac{c}{(a+c)(b+c)}
   \]
   We need to show that:
   \[
   \frac{c}{(a+c)(b+c)} \leq \frac{1}{20}
   \]
   This can be rewritten as:
   \[
   \frac{1}{20} \leq \frac{1}{20 + \frac{ab}{c}}
   \]
   Since \( a, b, c \geq 0 \), the inequality holds, proving the base case.

2. **Inductive Step: Assume the statement is true for \( n-1 \)**

   Suppose the statement holds for \( n-1 \). We need to show it holds for \( n \). Let \( A \) and \( B \) be subsets of \( P \) where \( P \) is the set of all \( n \)-tuples. Define:
   \[
   A = \bigcup_{i=1}^{20} A_i
   \]
   where \( A_i \) consists of all \( (x_1, x_2, \ldots, x_{n-1}) \) such that \( (x_1, x_2, \ldots, x_{n-1}, i) \in A \). By definition, \( |A| = \sum_{i=1}^{20} |A_i| \) and \( A_1 \supseteq A_2 \supseteq \cdots \supseteq A_{20} \).

   Similarly, define:
   \[
   B = \bigcup_{i=1}^{20} B_i
   \]
   where \( B_i \) consists of all \( (x_1, x_2, \ldots, x_{n-1}) \) such that \( (x_1, x_2, \ldots, x_{n-1}, i) \in B \). By definition, \( |B| = \sum_{i=1}^{20} |B_i| \) and \( B_1 \subseteq B_2 \subseteq \cdots \subseteq B_{20} \).

3. **Combining the Sets:**

   Using the inductive hypothesis:
   \[
   |A \cap B| = \sum_{i=1}^{20} |A_i \cap B_i| \leq \frac{1}{20^{n-1}} \sum_{i=1}^{20} |A_i| \cdot |B_i|
   \]
   Applying Chebyshev's rearrangement inequality:
   \[
   \sum_{i=1}^{20} |A_i| \cdot |B_i| \leq \frac{1}{20} \left( \sum_{i=1}^{20} |A_i| \right) \left( \sum_{i=1}^{20} |B_i| \right)
   \]
   Therefore:
   \[
   |A \cap B| \leq \frac{1}{20^{n-1}} \cdot \frac{1}{20} \left( \sum_{i=1}^{20} |A_i| \right) \left( \sum_{i=1}^{20} |B_i| \right) = \frac{1}{20^n} |A| \cdot |B|
   \]
   Hence:
   \[
   f(A, B) = \frac{|A \cap B|}{|A| \cdot |B|} \leq \frac{1}{20^n}
   \]

The final answer is \(\boxed{\frac{1}{20^{2012}}}\).