In each cell of $5 \times 5$ table there is one number from $1$ to $5$ such that every number occurs exactly once in every row and in every column. Number in one column is [i]good positioned[/i] if following holds:
 - In every row, every number which is left from [i]good positoned[/i] number is smaller than him, and every number which is right to him is greater than him, or vice versa.
 -  In every column, every number which is above from [i]good positoned[/i] number is smaller than him, and every number which is below to him is greater than him, or vice versa.
What is maximal number of good positioned numbers that can occur in this table?
1. **Understanding the Problem:**
   We are given a \(5 \times 5\) table where each cell contains a number from 1 to 5, and each number appears exactly once in every row and every column. A number is considered "good positioned" if:
   - In its row, all numbers to its left are either all smaller or all larger, and all numbers to its right are the opposite.
   - In its column, all numbers above it are either all smaller or all larger, and all numbers below it are the opposite.

2. **Constructing a Table with Good Positioned Numbers:**
   We need to construct a table where the maximum number of good positioned numbers is achieved. Consider the following table:
   \[
   \begin{array}{ccccc}
   1 & 4 & 2 & 3 & 5 \\
   3 & 5 & 1 & 4 & 2 \\
   2 & 1 & 3 & 5 & 4 \\
   4 & 2 & 5 & 1 & 3 \\
   5 & 3 & 4 & 2 & 1 \\
   \end{array}
   \]
   In this table, the numbers 1 and 5 in the corners and the number 3 in the center are all good positioned.

3. **Verifying the Good Positioned Numbers:**
   - **Corners (1 and 5):**
     - For 1 in the first row and first column, all numbers to its right and below are larger.
     - For 5 in the first row and fifth column, all numbers to its left and below are smaller.
     - For 1 in the fifth row and fifth column, all numbers to its left and above are larger.
     - For 5 in the fifth row and first column, all numbers to its right and above are smaller.
   - **Center (3):**
     - For 3 in the third row and third column, all numbers to its left and above are smaller, and all numbers to its right and below are larger.

4. **Proving the Maximum Number of Good Positioned Numbers:**
   - A number \(k\) can only be good if it is in either the \(k\)-th or \((6-k)\)-th column, and the \(k\)-th or \((6-k)\)-th row, because it needs \(k-1\) numbers above or below it, and \(k-1\) to the left or right.
   - This means that only the following numbers can be good:
     - The 1 in the first or fifth column.
     - The 2 in the second or fourth column.
     - The 3 in the third column.
     - The 4 in the second or fourth column.
     - The 5 in the first or fifth column.
   - We pair these possibilities:
     - The 1 in the first column and the 2 in the second column.
     - The 1 in the fifth column and the 2 in the fourth column.
     - The 3 in the third column.
     - The 4 in the second column and the 5 in the first column.
     - The 4 in the fourth column and the 5 in the fifth column.
   - Only one number in each pair can be good. For example, if the 2 in the second column is good, it must be in the second or fourth row and have a 1 to its left, which means the 1 in the first column cannot be in the first or fifth row, so it cannot be good.

5. **Conclusion:**
   - By applying similar arguments for the other pairs, we see that only one in each pair can be good.
   - Therefore, the maximum number of good positioned numbers is 5, which can be achieved as shown in the constructed table.

The final answer is \( \boxed{ 5 } \).