Find the sum of all the digits in the decimal representations of all the positive integers less than $1000.$
1. **Counting from 000 to 999:**
   - We include leading zeros and the number 0 itself, so we count from 000 to 999. This gives us a total of 1000 numbers.

2. **Sum of the units digits:**
   - The units digit cycles through 0 to 9 repeatedly. Each complete cycle (0 through 9) sums to:
     \[
     0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
     \]
   - There are 1000 numbers, so the units digit cycle repeats:
     \[
     \frac{1000}{10} = 100 \text{ times}
     \]
   - Therefore, the sum of the units digits is:
     \[
     45 \times 100 = 4500
     \]

3. **Sum of the tens digits:**
   - The tens digit also cycles through 0 to 9, but each digit appears 10 times in each complete cycle of 100 numbers (e.g., 00-09, 10-19, ..., 90-99).
   - Each complete cycle of 100 numbers sums to:
     \[
     10 \times (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 10 \times 45 = 450
     \]
   - There are 10 such cycles in 1000 numbers:
     \[
     \frac{1000}{100} = 10 \text{ times}
     \]
   - Therefore, the sum of the tens digits is:
     \[
     450 \times 10 = 4500
     \]

4. **Sum of the hundreds digits:**
   - The hundreds digit cycles through 0 to 9, but each digit appears 100 times in each complete cycle of 1000 numbers (e.g., 000-099, 100-199, ..., 900-999).
   - Each complete cycle of 1000 numbers sums to:
     \[
     100 \times (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 100 \times 45 = 4500
     \]

5. **Total sum of all digits:**
   - By symmetry, the sum of the units, tens, and hundreds digits are the same.
   - Therefore, the total sum of all digits is:
     \[
     4500 + 4500 + 4500 = 13500
     \]

The final answer is $\boxed{13500}$