Video solution:

Click to load, then click again to play

Written solution:

Call the integers from $1$ to $10$ inclusive the lower range, and call the integers from $11$ to $20$ inclusive the higher range.

Each of the $5$ distinct numbers chosen from the lower range blocks out the number in the higher range that is exactly $10$ more than itself. There are only $10$ numbers in the higher range, so there are only $10 - 5 = 5$ numbers not yet blocked.

We need to choose $5$ distinct numbers from the higher range, so the numbers chosen from the higher range are precisely those which are not yet blocked. They are each exactly $10$ more than a not-chosen number in the lower range.

So, the sum of the $5$ distinct numbers chosen from the higher range is exactly $5 \times 10 = 50$ more than the sum of the $5$ not-chosen numbers in the lower range.

The sum of all $10$ chosen numbers is therefore equal to $50$ plus the sum of all chosen and not-chosen numbers in the lower range $1 + 2 + \dots + 10.$

The sum of the numbers from $1$ to $10$ is $\frac{10 (10+1)}{2} = 55,$ so the answer is $50 + 55 = 105,$ or choice C.