A polygon on a coordinate grid is built of $ 2005$ dominoes $ 1 \times 2$. What is the smallest number of sides of an even length such a polygon can have?
1. **Understanding the Problem:**
   We are given a polygon built from 2005 dominoes, each of size \(1 \times 2\). We need to determine the smallest number of sides of even length that such a polygon can have.

2. **Initial Considerations:**
   Each domino covers two unit squares. Therefore, the total area covered by the dominoes is \(2005 \times 2 = 4010\) unit squares. 

3. **Chessboard Coloring Argument:**
   Consider a standard chessboard coloring of the coordinate grid. Each unit square is either black or white, and adjacent squares have different colors. Each domino covers one black and one white square.

4. **Vertices and Side Lengths:**
   Let \(P\) be the polygon formed by the dominoes. The vertices of \(P\) lie on the grid points. For each vertex \(p\) in \(P\), define \(f(p)\) as the difference between the number of black and white squares incident to \(p\). 

5. **Sum of \(f(p)\):**
   For vertices not on the boundary of \(P\), \(f(p) = 0\). For vertices on the boundary, \(f(p)\) can be \(-1\) or \(1\) depending on the orientation of the sides. Since the polygon has all odd sides, the sum of \(f(p)\) over all vertices must be non-zero, implying an imbalance between black and white squares.

6. **Green's Theorem Application:**
   Using Green's theorem, we can relate the sum of \(f(p)\) to the geometry of the polygon. Specifically, if all sides were odd, the sum would not balance, leading to a contradiction.

7. **Height Function Argument:**
   The height function argument states that for any cycle enclosing a region tiled by dominoes, the sum of the height function over the vertices of the cycle must vanish. If all sides were odd, this sum would not vanish, leading to another contradiction.

8. **Conclusion:**
   Since having all sides of odd length leads to contradictions, the polygon must have at least some sides of even length. The smallest number of sides of even length that balances the conditions is 2. This is because having exactly two sides of even length can balance the sum of \(f(p)\) while maintaining the tiling by dominoes.

The final answer is \(\boxed{2}\).