Let $n$ be a positive integer. Find the greatest possible integer $m$, in terms of $n$, with the following property: a table with $m$ rows and $n$ columns can be filled with real numbers in such a manner that for any two different rows $\left[ {{a_1},{a_2},\ldots,{a_n}}\right]$ and $\left[ {{b_1},{b_2},\ldots,{b_n}} \right]$ the following holds: \[\max\left( {\left| {{a_1} - {b_1}} \right|,\left| {{a_2} - {b_2}} \right|,...,\left| {{a_n} - {b_n}} \right|} \right) = 1\]
[i]Poland (Tomasz Kobos)[/i]
1. **Understanding the Problem:**
   We need to find the maximum number of rows \( m \) in a table with \( n \) columns such that for any two different rows \(\left[ a_1, a_2, \ldots, a_n \right]\) and \(\left[ b_1, b_2, \ldots, b_n \right]\), the maximum absolute difference between corresponding elements is exactly 1:
   \[
   \max\left( \left| a_1 - b_1 \right|, \left| a_2 - b_2 \right|, \ldots, \left| a_n - b_n \right| \right) = 1
   \]

2. **Constructing the Table:**
   Consider the binary vectors of length \( n \). There are \( 2^n \) such vectors. Each element of these vectors can be either 0 or 1. Let's denote these vectors as rows in our table.

3. **Verifying the Condition:**
   For any two different binary vectors \(\left[ a_1, a_2, \ldots, a_n \right]\) and \(\left[ b_1, b_2, \ldots, b_n \right]\), since they are different, there exists at least one position \( i \) where \( a_i \neq b_i \). In binary vectors, this difference is exactly 1. Therefore:
   \[
   \max\left( \left| a_1 - b_1 \right|, \left| a_2 - b_2 \right|, \ldots, \left| a_n - b_n \right| \right) = 1
   \]
   This satisfies the given condition.

4. **Maximizing \( m \):**
   We have shown that with \( 2^n \) binary vectors, the condition is satisfied. Now, we need to prove that \( m \) cannot be greater than \( 2^n \).

5. **Proof by Contradiction:**
   Assume \( m > 2^n \). This means we have more than \( 2^n \) rows. By the pigeonhole principle, at least two rows must be identical in some columns, which would violate the condition that the maximum absolute difference between corresponding elements is exactly 1. Hence, \( m \) cannot be greater than \( 2^n \).

Conclusion:
The greatest possible integer \( m \) is \( 2^n \).

The final answer is \( \boxed{2^n} \).