How many $5-$digit positive numbers which contain only odd numbers are there such that there is at least one pair of consecutive digits whose sum is $10$?

$ 
\textbf{(A)}\ 3125
\qquad\textbf{(B)}\ 2500
\qquad\textbf{(C)}\ 1845
\qquad\textbf{(D)}\ 1190
\qquad\textbf{(E)}\ \text{None of the preceding}
$
1. We need to find the number of 5-digit positive numbers that contain only odd digits and have at least one pair of consecutive digits whose sum is 10.
2. The odd digits are {1, 3, 5, 7, 9}. We need to count the total number of 5-digit numbers that can be formed using these digits.
3. Each digit in the 5-digit number can be any of the 5 odd digits. Therefore, the total number of 5-digit numbers is:
   \[
   5^5 = 3125
   \]
4. Next, we count the number of 5-digit numbers that do not have any pair of consecutive digits whose sum is 10. 
5. For any given odd digit, there is exactly one other odd digit that makes the sum 10:
   - 1 pairs with 9
   - 3 pairs with 7
   - 5 pairs with 5
   - 7 pairs with 3
   - 9 pairs with 1
6. To ensure no two consecutive digits sum to 10, we have 5 choices for the first digit, and for each subsequent digit, we have 4 choices (excluding the digit that would sum to 10 with the previous digit).
7. Therefore, the number of 5-digit numbers without any pair of consecutive digits summing to 10 is:
   \[
   5 \cdot 4^4 = 5 \cdot 256 = 1280
   \]
8. To find the number of 5-digit numbers that have at least one pair of consecutive digits whose sum is 10, we subtract the number of 5-digit numbers without such pairs from the total number of 5-digit numbers:
   \[
   5^5 - 5 \cdot 4^4 = 3125 - 1280 = 1845
   \]

The final answer is $\boxed{1845}$