To form a two-digit odd number, we need to consider the following:

1. First Digit (Tens Place): This digit cannot be zero, as we need a two-digit number. Thus, the first digit can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ or $9$. This gives us 9 possible choices for the first digit.

2. Second Digit (Units Place): Since the number needs to be odd, the units place must be an odd digit. The odd digits from $0$ to $9$ are $1, 3, 5, 7,$ and $9$, giving us 5 possible choices for the second digit.

Now, we can calculate the total number of two-digit odd numbers by multiplying the number of choices for each digit:

$9 \times 5 = 45$

Conclusion:

There are 45 possible two-digit odd numbers that can be formed with replacement from the digits $0$ through $9$.

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Would you like further details on this solution or have any questions?

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Here are some related questions that could deepen understanding:

1. How many two-digit even numbers can be formed with the same digits?
2. What if the digits were formed without replacement? How would that change the number of odd numbers?
3. How many two-digit natural numbers in total can be formed from the digits 0 through 9?
4. What if only prime numbers were considered instead of odd numbers?
5. How many three-digit odd numbers can be formed with replacement using these digits?

Tip: Always check for restrictions (like "odd" or "even") when forming numbers, as these influence the digit choices in specific places.