Solution:

We first arrange the numbers $670, 671, \ldots, 2007$ into 223 rows and 6 columns in the following way:

Let $C_{i}$ represent the set containing the numbers in the $i$$\text{th}$ row of the above arrangement. It is easy to check that the numbers in each $C_{i}$ add up to a constant sum.

We now need to arrange the numbers $1, 2, \ldots, 669$ into 223 rows and 3 columns in such a way that the sum of the numbers in each row is the same for all the rows:

Note that the sum of the numbers in the first and second columns of one row is different from the sum of the numbers in the first and second columns of another row. Since we expect that the sum of the numbers in each row is $(669)(670) \div (2)(223) = 1005$, we choose the number in the third column of a row to be the difference between 1005 and the sum of its numbers in the first and second columns. Thus, the numbers in the third column are distinct. Since the sum of the numbers in the first and second columns in each row ranges from 336 to 558, we expect that the numbers in the third column are the numbers from $1005 - 558 = 447$ to $1005 - 336 = 669$.

Let $B_{i}$ be the set containing the numbers in the $i$$\text{th}$ row of the above arrangement.

The desired decomposition of the set $\{1,2, \ldots, 2007\}$ is $A_{i} = B_{i} \cup C_{i}$, $i = 1, 2, \ldots, 223$.