1. Understanding the problem: We need to fill a $3 \times 6$ grid with the numbers from the list $\{1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9\}$ such that:
- Both the first three columns and the last three columns contain the integers 1 through 9.
- No numbers appear more than once in a given row.
- We need to find the largest integer $k$ such that $2^k$ divides the number of ways to fill the grid.

2. **Filling the first $3 \times 3$ grid**: The first $3 \times 3$ grid must contain the numbers 1 through 9. There are $9!$ ways to arrange these numbers in the first $3 \times 3$ grid.

3. **Filling the second $3 \times 3$ grid**: The second $3 \times 3$ grid must also contain the numbers 1 through 9. However, the numbers in each row of the second $3 \times 3$ grid must be different from the numbers in the corresponding row of the first $3 \times 3$ grid.

4. Counting the arrangements:
- For each row in the first $3 \times 3$ grid, there are $6$ remaining numbers that can be placed in the corresponding row of the second $3 \times 3$ grid.
- The number of ways to arrange these $6$ numbers in each row is $6!$.
- Since there are $3$ rows, the total number of ways to arrange the second $3 \times 3$ grid is $(6!)^3$.

5. Combining the arrangements: The total number of ways to fill the entire $3 \times 6$ grid is:
$$9! \times (6!)^3$$

6. **Finding the largest power of 2 that divides $N$**:
- The prime factorization of $9!$ is:
$$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 = 2^7 \times 3^4 \times 5 \times 7$$
- The prime factorization of $6!$ is:
$$6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 2^4 \times 3^2 \times 5$$
- Therefore, the prime factorization of $(6!)^3$ is:
$$(6!)^3 = (2^4 \times 3^2 \times 5)^3 = 2^{12} \times 3^6 \times 5^3$$
- Combining these, the prime factorization of $9! \times (6!)^3$ is:
$$9! \times (6!)^3 = 2^7 \times 3^4 \times 5 \times 7 \times 2^{12} \times 3^6 \times 5^3 = 2^{19} \times 3^{10} \times 5^4 \times 7$$

7. Conclusion: The largest power of 2 that divides $N$ is $2^{19}$, so $k = 19$.

The final answer is $\boxed{19}$.