The given question asks us to find the sum of all numbers greater than 1000 that can be formed using the digits 1, 3, 5, and 7, with no digit being repeated in any number. Let's solve this problem step by step.

Step 1: Determine the total number of digits
Since we can form numbers greater than 1000, the thousands place will always be occupied by one of the four digits: 1, 3, 5, or 7. The units, tens, and hundreds places can be filled with any of the remaining three digits. Therefore, the total number of digits is 4 * 3 * 2 * 1 = 24.

Step 2: Calculate the sum of each digit at each place
To find the sum of each digit at each place, we need to consider all the possible combinations of the four digits.

For the thousands place, each digit (1, 3, 5, or 7) will occur 6 times because there are 6 possible combinations of the remaining three digits for each digit at the thousands place. So, the sum of the digits at the thousands place will be (1+3+5+7) * 6 = 96.

Similarly, for the hundreds, tens, and units places, each digit will occur 6 times. So, the sum of the digits at each of these places will also be (1+3+5+7) * 6 = 96.

Step 3: Calculate the sum of all the numbers
To find the sum of all the numbers, we need to multiply the sum of each digit at each place by the place value (1000, 100, 10, and 1) and then add them together.

Sum = (96 * 1000) + (96 * 100) + (96 * 10) + (96 * 1)
= 96,000 + 9,600 + 960 + 96
= 106,656

Therefore, the sum of all numbers greater than 1000 formed using the digits 1, 3, 5, and 7, with no digit being repeated in any number, is 106,656. Hence, the correct answer is option C.