I suddenly was in a number-crunching mood, so…
I listed all fifty states by the number of their electoral votes. I also found a chart of the states’ total populations (as of 2001).
Then I added up the high-EV states until I got to 268 (I had to leap out of order at the end), i.e., just barely losing. I further pretended that each state’s popular vote for its winning candidate was 50% plus 1 (the lowest possible). I further tweaked the undemocratic nature of this exercise by assuming that voter turnout in low-EV states was very low (20% of eligible voters) and very high in high-EV states (80% of eligible voters). I assumed the percentage of eligible voters was the same in each state.
What are the results? Let’s say Candidate A wins the following high-EV states (with, as I said, 50%-plus-one votes, with 80% voter turnout):
California, Texas, New York, Florida, Illinois, Pennsylvania, Ohio, Michigan, Georgia, North Carolina, Massachusetts
Candidate B wins every other state, with 50%-plus-one votes and 20% voter turnout.
Who wins the electoral vote? Candidate B, with 270 electoral votes.
Who loses the electoral vote? Candidate A, with 268 electoral votes…and 84.3% of the popular vote.
Of course, such a concatenation of just-barelies is extremely unlikely…but the point is that such a result is theoretically possible. Even with a more equitable assumption (that every state has the same percentage of voter turnout), our losing candidate wins 57.3% of the popular vote.