Longest losing streak for a county.

[E-Dawg]

The longest current losing streak is now Kenedy County, Texas, the only Romney-Clinton-Trump county.

The next longest current losing streak would be the 14 other Clinton-Trump counties, which are:
Val Verde, TX
Reeves, TX
Frio, TX
La Salle, TX
Zapata, TX
Kennedy, TX
Kleberg, TX
Jim Wells, TX
Burke, GA
Clarendon, SC
Dillon, SC
Scotland, NC
Lorain, OH
Mahoning, OH
Alamosa, CO

[E-Dawg]

Update: I revisited the 1860-1880 data and I think I have the definitive answer. Hopefully we can lay this question to rest.

By my count, 199 counties voted for a losing candidate in all 6 elections.

Twenty-odd of those counties, mostly in Kentucky, voted for Fillmore in 1856. (Only a handful voted for Frémont.) Kentucky being the home of Henry Clay and a Whig stronghold at the time, 16 such counties also voted for Scott in 1852, extending their streak to 8. However, their streaks all end there as Zach Taylor won and carried all of these counties in 1848.

Only two of the original 199 counties voted for James Blaine in 1884 – Greene, NY and Livingston, MO. Remarkably, both flipped back to Cleveland in 1888, extending their streaks to 8.

Greene then voted for Harrison in 1892, making it 9. That streak ended when Greene voted for McKinley in 1896.

So there's your answer. Greene County, NY (which is a lovely place) picked losers in nine consecutive elections: 1860–92.

It is the nation's only Douglas*–McClellan–Seymour–Greeley–Tilden–Hancock–Blaine–Cleveland–Harrison county.

So Greene County, New York voted for the loser every election from 1860-1892, giving it a losing streak of 9 elections.

[E-Dawg]

It's not as big as a losing streak as the ones above, but I found it interesting that Butte County CA had a 5 election losing streak from 1880 through 1896, voting Hancock-Blaine-Cleveland-Harrison-Bryan. This was a particularly cool streak since Butte County and the nation switched parties every election during this period, just in the exact opposite way.