 You may have found an idiot proofing error, but in 99.9 per cent of the  voltage drop problems a user that is familiar with electrical work can use  the program to find practical useful answers. The program is not written  for idiots; it is written for electricians.  hint: electrician.com
I'll believe physicists and engineers before I'll believe an electrician.
 If the answer is negative, then obviously all the volts are dropped in the  conductor. Any 1st year electrician apprentice should know that. But
Really? You think there will be ZERO VOLTS at the load? You really have no concept of physics, do you.
According to my tables, the 5280 foot run of 18 AWG copper wire, at a rating of 6.3851 ohms per 1000 feet, will have 33.713328 ohms for the full 5280 feet, and 67.426656 ohms for the full loop. Adding in the 12 ohms of load makes the total ohmage of 79.426656. So, at 120 volts that's a current of
1.5108278 amps. That puts a voltage difference of 18.1299336 volts across the load, and 101.8700664 volts across the wiring. The total watts in the load is 27.3912077 (divided across the 20 bulbs means each dissipates only
1.36956 watts). The remaining 153.9081283 watts is dissipated in the wires at 0.0145746 watts per foot per conductor.
 more than that, the ampacity of No. 18 is not 10 amperes. Voltage drop  alone does not solve all the problem, the ampacity of a conductor is solved  using different equations and Tables. Additionally, for No. 18 NEC  Section 726.23 requires overcurrent protection no larger than 7 amperes.  Also, the advanced voltage drop calculator finds the approximate temperature  of a conductor and finds the voltage drop for that temperature. Read the  article
Did you mean 725.23?
I suggested looking at table 310.16 under the 90 degree insulation column for copper and note the ampacity for 18 AWG wire there is 14. That's for wiring in raceways or buried, up to 3 conductors. Now have a look at table
310.17 and see what the ampacity for 18 AWG wire in free air is 18.
For the purpose of calculating voltage drop, the ampacity is not really relevant. The temperature would be, but the temperature coefficient does not alter the above values radically.
And finally, with a total resistance of 79.426656 ohms at 120 volts, the current is only 1.5108278 amps, anyway. That's well below virtually every rating for 18 AWG wire. You might not have much light, but the wire is definitely not going to burn up.

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 for some enlightenment.   You appear to be more into electronics and ham radio than electrical work.  Perhaps you should stay away from the electrician page until you serve an  8,000 hour apprenticeship and an additional 8,000 hours as a Journeyman  electrician and then get your electrical administrators license then you  might be able to use electrical calculators without evaluating them as a  technician.
If electricians use calculators or formulas that result in violations of physics, then I sure as hell don't want to hire them to wire anything of mine. I'll play it safe and wire my own.