The integers from $1$ to $1000000$ are divided into two groups consisting of numbers with odd or even sums of digits respectively. Which group contains more numbers?
1. **Define the sets:**
   Let \( E \) be the set of integers from 1 to 1,000,000 with an even sum of digits, and \( O \) be the set of integers from 1 to 1,000,000 with an odd sum of digits.

2. **Construct a bijection:**
   Consider any number \( n \) in the range \( 2 \leq n \leq 999999 \). We can create a bijection between the sets \( E \) and \( O \) by modifying the last digit of \( n \):
   - If the last digit of \( n \) is even, add 1 to it.
   - If the last digit of \( n \) is odd, subtract 1 from it.

   This operation changes the parity of the sum of the digits of \( n \), thus mapping an element from \( E \) to \( O \) and vice versa.

3. **Count the elements:**
   The bijection shows that for every number \( n \) in the range \( 2 \leq n \leq 999999 \), there is a corresponding number in the other set. Therefore, the sets \( E \) and \( O \) have the same number of elements in this range.

4. **Consider the boundary elements:**
   The numbers 1 and 1,000,000 are not included in the bijection:
   - The sum of the digits of 1 is 1, which is odd, so 1 belongs to \( O \).
   - The sum of the digits of 1,000,000 is 1, which is odd, so 1,000,000 belongs to \( O \).

5. **Conclusion:**
   Since the sets \( E \) and \( O \) have the same number of elements in the range \( 2 \leq n \leq 999999 \), and both 1 and 1,000,000 belong to \( O \), the set \( O \) has two more elements than \( E \).

The final answer is \( \boxed{ O } \) contains more numbers.