The numbers from 1 to $ 2013^2 $ are written row by row into a table consisting of $ 2013 \times 2013 $ cells. Afterwards, all columns and all rows containing at least one of the perfect squares $ 1, 4, 9, \cdots, 2013^2 $ are simultaneously deleted.
How many cells remain?
1. **Identify the perfect squares in the range:**
   The perfect squares in the range from \(1\) to \(2013^2\) are \(1^2, 2^2, 3^2, \ldots, 2013^2\). There are \(2013\) perfect squares in total.

2. **Determine the number of rows and columns containing at least one perfect square:**
   - Since the table is \(2013 \times 2013\), each row and each column will contain exactly one perfect square. This is because the perfect squares are distributed sequentially from \(1\) to \(2013^2\).
   - Therefore, each of the \(2013\) rows and each of the \(2013\) columns will contain at least one perfect square.

3. **Calculate the number of cells removed:**
   - If we delete all rows containing at least one perfect square, we delete \(2013\) rows.
   - If we delete all columns containing at least one perfect square, we delete \(2013\) columns.
   - However, the intersection of these rows and columns (i.e., the cells where both a row and a column containing a perfect square intersect) will be counted twice. There are \(2013\) such intersections (one for each perfect square).

4. **Calculate the total number of cells removed:**
   \[
   \text{Total cells removed} = 2013 \times 2013 + 2013 \times 2013 - 2013 = 2 \times 2013^2 - 2013
   \]

5. **Calculate the number of remaining cells:**
   - The total number of cells in the table is \(2013 \times 2013\).
   - The number of remaining cells is:
   \[
   2013^2 - (2 \times 2013^2 - 2013) = 2013^2 - 2 \times 2013^2 + 2013 = -2013^2 + 2013
   \]

6. **Simplify the expression:**
   \[
   -2013^2 + 2013 = 2013(1 - 2013) = 2013 \times (-2012) = -2013 \times 2012
   \]

Since the number of cells cannot be negative, we need to re-evaluate the logic. The correct approach should be:

1. **Re-evaluate the number of rows and columns containing perfect squares:**
   - Each row and each column will contain at least one perfect square, so we delete all \(2013\) rows and all \(2013\) columns.

2. **Correct calculation of remaining cells:**
   - Total cells in the table: \(2013 \times 2013\)
   - Cells removed due to rows: \(2013 \times 2013\)
   - Cells removed due to columns: \(2013 \times 2013\)
   - Cells counted twice (intersection of rows and columns): \(2013\)

   Therefore, the correct number of remaining cells is:
   \[
   2013^2 - (2013 \times 2013 + 2013 \times 2013 - 2013) = 2013^2 - (2 \times 2013^2 - 2013) = 2013^2 - 2 \times 2013^2 + 2013 = -2013^2 + 2013
   \]

   This again leads to a negative number, indicating a mistake in the logic. The correct approach should be:

   - Total cells: \(2013 \times 2013\)
   - Cells removed: \(2013 \times 2013\) (since each row and column is deleted once)

   Therefore, the number of remaining cells is:
   \[
   2013^2 - 2013^2 = 0
   \]

The final answer is \(\boxed{0}\)