I think
50
because
from 1 to 100,
there are 50 odd no. and 50 even no.

Introduction:
To find the number of 4-digit odd numbers without repetition of digits, we need to consider the possible combinations of digits and determine how many of them are odd.

Understanding the Problem:
We are given the following conditions:
1. The number should be odd, which means the last digit must be odd.
2. Repetition of digits is not allowed, so each digit in the number should be unique.
3. We need to find the total count of such numbers.

Step-by-Step Solution:
To solve this problem, we can break it down into smaller steps:

1. Determine the possible choices for each digit:
- For the first digit, we cannot choose zero as it will make it a 3-digit number. So we have 9 choices (1-9).
- For the second digit, we have 9 remaining choices (0-9 excluding the digit chosen for the first position).
- Similarly, for the third digit, we have 8 remaining choices.
- For the last digit (the units place), we have 5 choices (1, 3, 5, 7, 9) as it needs to be odd.

2. Count the possible combinations:
Now, let's calculate the total count of 4-digit odd numbers without repetition using the choices determined in the previous step.

- The number of choices for the first digit is 9.
- The number of choices for the second digit is 9.
- The number of choices for the third digit is 8.
- The number of choices for the last digit is 5.

Therefore, the total count of 4-digit odd numbers without repetition can be calculated as:
9 * 9 * 8 * 5 = 3240

Conclusion:
There are a total of 3240 four-digit odd numbers without repetition.