Harcourt is a city in Webster County, Iowa, United States.

[3] It was named for William Vernon Harcourt, a British statesman.

[5] According to the United States Census Bureau, the city has a total area of 1.00 square mile (2.59 km2), all land.

There were 138 housing units at an average density of 136.5 per square mile (52.7/km2).

Hispanic or Latino persons of any race comprised 4.5% of the population.

There were 146 housing units at an average density of 146.0 per square mile (56.4/km2).

29.0% of all households were made up of individuals, and 9.9% had someone living alone who was 65 years of age or older.

There were 149 housing units at an average density of 259.7 per square mile (100.3/km2).

35.0% of all households were made up of individuals, and 18.2% had someone living alone who was 65 years of age or older.

None of the families and 3.3% of the population were living below the poverty line, including no under eighteens and 4.5% of those over 64.