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Milford from the air, looking northeast.

Milford is a town in Decatur County, Indiana, United States. The population was 121 at the 2000 census. Milford has the dubious distinction of having the smallest population of any incorporated community in the State of Indiana.

Milford is located at 39°21′1″N, 85°37′10″W (39.350176, -85.619429)^GR1.

According to the United States Census Bureau, the town has a total area of 0.2 km² (0.1 mi²), all land.

As of the census^GR2 of 2000, there were 121 people, 48 households, and 37 families residing in the town. The population density was 519.1/km² (1,349.1/mi²). There were 54 housing units at an average density of 231.7/km² (602.1/mi²). The racial makeup of the town was 100.00% White.

There were 48 households out of which 29.2% had children under the age of 18 living with them, 60.4% were married couples living together, 8.3% had a female householder with no husband present, and 22.9% were non-families. 20.8% of all households were made up of individuals and 6.3% had someone living alone who was 65 years of age or older. The average household size was 2.52 and the average family size was 2.89.

In the town the population was spread out with 25.6% under the age of 18, 12.4% from 18 to 24, 24.0% from 25 to 44, 27.3% from 45 to 64, and 10.7% who were 65 years of age or older. The median age was 37 years. For every 100 females there were 105.1 males. For every 100 females age 18 and over, there were 95.7 males.

The median income for a household in the town was $30,781, and the median income for a family was $31,406. Males had a median income of $31,250 versus $19,167 for females. The per capita income for the town was $12,506. There were 12.2% of families and 15.9% of the population living below the poverty line, including 3.4% of under eighteens and 11.5% of those over 64.

• Maps and aerial photos Coordinates: 39.350176° -85.619429°
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