What is the maximum number of coins can be arranged in cells of the table $n \times n$ (each cell is not more of the one coin) so that any coin was not simultaneously below and to the right than any other?
To solve the problem of finding the maximum number of coins that can be arranged in an \( n \times n \) table such that no coin is simultaneously below and to the right of any other coin, we need to consider the constraints and possible configurations.

1. **Understanding the Constraints:**
   - No two coins should be in the same row or column.
   - No coin should be below and to the right of any other coin.

2. **Case 1: No two coins in the same row or column:**
   - If no two coins are in the same row or column, the maximum number of coins that can be placed is \( n \). This is because each row and each column can contain at most one coin. One possible arrangement is placing the coins on the main diagonal of the \( n \times n \) table.

3. **Case 2: No coin below and to the right of any other coin:**
   - To ensure that no coin is below and to the right of any other coin, we can place the coins in a "chessboard" pattern. This means placing coins in alternating cells starting from \((1,1)\).

4. **Calculating the Maximum Number of Coins:**
   - For an even \( n \):
     \[
     f(n) = \frac{n^2}{2}
     \]
     This is because half of the cells in an \( n \times n \) table will be occupied in a chessboard pattern.

   - For an odd \( n \):
     \[
     f(n) = 2\left\lfloor \frac{n}{2} \right\rfloor^2 + 2\left\lfloor \frac{n}{2} \right\rfloor + 1
     \]
     This formula accounts for the additional row and column that appear when \( n \) is odd.

5. **Verification:**
   - For \( n = 2 \):
     \[
     f(2) = \frac{2^2}{2} = 2
     \]
     The arrangement would be:
     \[
     \begin{array}{cc}
     \bullet & \\
     & \bullet \\
     \end{array}
     \]

   - For \( n = 3 \):
     \[
     f(3) = 2\left\lfloor \frac{3}{2} \right\rfloor^2 + 2\left\lfloor \frac{3}{2} \right\rfloor + 1 = 2(1)^2 + 2(1) + 1 = 5
     \]
     The arrangement would be:
     \[
     \begin{array}{ccc}
     \bullet & & \bullet \\
     & \bullet & \\
     \bullet & & \bullet \\
     \end{array}
     \]

Conclusion:
The maximum number of coins that can be arranged in an \( n \times n \) table such that no coin is simultaneously below and to the right of any other coin is given by:
\[
f(n) = 
\begin{cases} 
\frac{n^2}{2} & \text{if } n \text{ is even} \\
2\left\lfloor \frac{n}{2} \right\rfloor^2 + 2\left\lfloor \frac{n}{2} \right\rfloor + 1 & \text{if } n \text{ is odd}
\end{cases}
\]

The final answer is \( \boxed{ f(n) = \frac{n^2}{2} } \) if \( n \) is even, and \( f(n) = 2\left\lfloor \frac{n}{2} \right\rfloor^2 + 2\left\lfloor \frac{n}{2} \right\rfloor + 1 \) if \( n \) is odd.