Numbers $0, 1$ and $2$ are placed in a table $2005 \times 2006$ so that total sums of the numbers in each row and in each column are factors of $3$. Find the maximal possible number of $1$'s that can be placed in the table.

[i](6 points)[/i]
1. **Understanding the Problem:**
   We need to place the numbers \(0\), \(1\), and \(2\) in a \(2005 \times 2006\) table such that the sum of the numbers in each row is congruent to \(1 \pmod{3}\) and the sum of the numbers in each column is congruent to \(2 \pmod{3}\). We aim to find the maximal possible number of \(1\)'s that can be placed in the table.

2. **Setting Up the Equations:**
   Let \(m\) be the number of \(1\)'s and \(n\) be the number of \(2\)'s in the table. Each row must have a sum of \(1 \pmod{3}\), and each column must have a sum of \(2 \pmod{3}\).

   - For each row to sum to \(1 \pmod{3}\), it can either have one \(1\) or two \(2\)'s.
   - For each column to sum to \(2 \pmod{3}\), it can either have two \(1\)'s or one \(2\).

3. **Formulating the Constraints:**
   - The total number of \(1\)'s and \(2\)'s must satisfy the row condition:
     \[
     m + \frac{n}{2} \geq 2005
     \]
   - The total number of \(1\)'s and \(2\)'s must satisfy the column condition:
     \[
     n + \frac{m}{2} \geq 2006
     \]

4. **Summing the Constraints:**
   Adding the two inequalities:
   \[
   m + \frac{n}{2} + n + \frac{m}{2} \geq 2005 + 2006
   \]
   Simplifying, we get:
   \[
   \frac{3m}{2} + \frac{3n}{2} \geq 4011
   \]
   \[
   m + n \geq \frac{2}{3} \times 4011 = 2674
   \]

5. **Maximizing the Number of \(1\)'s:**
   To maximize the number of \(1\)'s, we need to minimize the number of \(2\)'s. From the equality \(m + n = 2674\), we can express \(n\) in terms of \(m\):
   \[
   n = 2674 - m
   \]

6. **Ensuring the Constraints are Met:**
   Substituting \(n = 2674 - m\) into the original constraints:
   \[
   m + \frac{2674 - m}{2} \geq 2005
   \]
   \[
   m + 1337 - \frac{m}{2} \geq 2005
   \]
   \[
   \frac{m}{2} + 1337 \geq 2005
   \]
   \[
   \frac{m}{2} \geq 668
   \]
   \[
   m \geq 1336
   \]

   Similarly, for the column constraint:
   \[
   2674 - m + \frac{m}{2} \geq 2006
   \]
   \[
   2674 - \frac{m}{2} \geq 2006
   \]
   \[
   2674 - 2006 \geq \frac{m}{2}
   \]
   \[
   668 \geq \frac{m}{2}
   \]
   \[
   1336 \geq m
   \]

   Therefore, \(m = 1336\) is the maximum number of \(1\)'s that can be placed in the table.

7. **Calculating the Total Number of \(1\)'s:**
   The total number of \(1\)'s is:
   \[
   m = 1336
   \]

The final answer is \(\boxed{1336}\)