Let $m$, $n$ be two positive integers greater than 3. Consider the table of size $m\times n$ ($m$ rows and $n$ columns)  formed with unit squares. We are putting marbles into unit squares of the table following the instructions:

$-$ each time put 4 marbles into 4 unit squares (1 marble per square) such that the 4 unit squares formes one of the followings 4 pictures (click [url=http://www.mathlinks.ro/Forum/download.php?id=4425]here[/url] to view the pictures). 

In each of the following cases, answer with justification to the following question: Is it possible that after a finite number of steps we can set the marbles into all of the unit squares such that the numbers of marbles in each unit square is the same? 

a) $m=2004$, $n=2006$;

b) $m=2005$, $n=2006$.
### Part (a): \( m = 2004 \), \( n = 2006 \)

1. **Divide the Table into Smaller Rectangles**:
   - The table is of size \( 2004 \times 2006 \).
   - We can divide this table into smaller rectangles of size \( 2 \times 4 \).

2. **Filling Each \( 2 \times 4 \) Rectangle**:
   - Each \( 2 \times 4 \) rectangle can be filled with marbles such that each unit square within the rectangle gets exactly one marble.
   - This is possible because the given patterns allow us to place one marble in each square of a \( 2 \times 4 \) rectangle.

3. **Covering the Entire Table**:
   - Since \( 2004 \) and \( 2006 \) are both even, the entire \( 2004 \times 2006 \) table can be completely covered by \( 2 \times 4 \) rectangles without any gaps or overlaps.
   - Therefore, it is possible to place one marble in each unit square of the \( 2004 \times 2006 \) table.

### Conclusion for Part (a):
\[
\boxed{\text{True}}
\]

### Part (b): \( m = 2005 \), \( n = 2006 \)

1. **Coloring the Table**:
   - Color the rows of the table alternately black and white, starting with white.
   - This means that the first row is white, the second row is black, and so on.

2. **Counting Black and White Squares**:
   - Since \( m = 2005 \) (odd) and \( n = 2006 \) (even), there will be \( 1003 \) black rows and \( 1002 \) white rows.
   - Each row has \( 2006 \) squares, so the number of black squares \( B \) and white squares \( W \) are:
     \[
     B = 1003 \times 2006, \quad W = 1002 \times 2006
     \]

3. **Invariance of \( B - W \)**:
   - Each operation (placing 4 marbles in a specific pattern) affects an equal number of black and white squares.
   - Therefore, the difference \( B - W \) remains constant after each operation.
   - Initially, \( B = W = 0 \), so \( B - W = 0 \).

4. **Final Configuration**:
   - To have the same number of marbles in each square, the difference \( B - W \) must be zero.
   - However, since there are \( 2006 \) more white squares than black squares, the difference \( B - W \) would be \( 2006 \times c \) where \( c \) is the number of marbles in each square.
   - This leads to a contradiction because \( B - W \) cannot be zero if \( c \neq 0 \).

### Conclusion for Part (b):
\[
\boxed{\text{False}}
\]