Board with dimesions $2018 \times 2018$ is divided in unit cells $1 \times 1$. In some cells of board are placed black chips and in some white chips (in every cell maximum is one chip). Firstly we remove all black chips from columns which contain white chips, and then we remove all white chips from rows which contain black chips. If $W$ is number of remaining white chips, and $B$ number of remaining black chips on board and $A=min\{W,B\}$, determine maximum of $A$
1. **Initial Setup and Problem Understanding**:
   - We have a \(2018 \times 2018\) board divided into \(1 \times 1\) unit cells.
   - Each cell can contain either a black chip or a white chip, but not both.
   - We perform two operations:
     1. Remove all black chips from columns that contain white chips.
     2. Remove all white chips from rows that contain black chips.
   - We need to determine the maximum value of \(A = \min\{W, B\}\), where \(W\) is the number of remaining white chips and \(B\) is the number of remaining black chips.

2. **End Configuration Analysis**:
   - After the operations, no row with a black chip can have a white chip, and no column with a white chip can have a black chip.
   - To maximize \(A\), we need to balance the number of remaining white and black chips.

3. **Optimal Placement Strategy**:
   - Suppose we place white chips in \(x\) distinct columns and \(y\) distinct rows.
   - The remaining black chips should be placed in the remaining \(2018 - x\) columns and \(2018 - y\) rows.
   - The goal is to maximize \(A = \min\{xy, (2018 - x)(2018 - y)\}\).

4. **Symmetry and Balance**:
   - To maximize the minimum of the two products, we should balance the number of rows and columns for both colors.
   - Let \(x = y = 1009\). This means we place white chips in a \(1009 \times 1009\) subgrid and black chips in the remaining \(1009 \times 1009\) subgrid.

5. **Calculation of Chips**:
   - Number of white chips: \(1009 \times 1009 = 1009^2\).
   - Number of black chips: \(1009 \times 1009 = 1009^2\).
   - Therefore, \(A = \min\{1009^2, 1009^2\} = 1009^2\).

6. **Verification of Maximum**:
   - Suppose there is a configuration with more than \(1009^2\) chips for one color. Then the other color must have fewer than \(1009^2\) chips, contradicting the balance.
   - If \(x\) and \(y\) are both less than 1009, or both greater than 1009, the product \(xy\) or \((2018 - x)(2018 - y)\) will be less than \(1009^2\).
   - Hence, the configuration \(x = y = 1009\) is optimal.

The final answer is \(\boxed{1018081}\).