If it is not already there, move the black bar at the bottom of the graph up so that it crosses the Y axis at 1. The bar should go right through the red circle. Notice the numerical indicator of the black bar immediately to its right.

The deviation of the red circle from the bar is 0, so you won't see a red rectangle on the right-hand portion of the graph. The line between the bar and the blue circle is the deviation of the circle from the bar. It has a length of 1. Notice that the height of the blue rectangle is 1.

The green line has a length of 2 and the height of the green rectangle is also 2. The total height of the rectangle is the sum of all the line lengths: 0 + 1 + 2 + 3 + 4 = 10. This height is the sum of the absolute deviations from the bar. It is marked below the rectangle.

Your goal is to find the placement of the bar that gives you the shortest rectangle. This will be the value that minimizes the sum of the rectangles. Move the bar up and down until you think you have found this value. Then, to make sure you are correct, click on the "OK" button at the bottom of the graph. This will move the black bar to the correct location. If nothing changes, you found the correct location on your own.

Now, change the value of the green circle from 3 to somewhere between 2 and 3. You move the circle by clicking on it and dragging it. Notice that the value of the point is shown at the bottom of the graph in green.

Next, find the value that minimizes the sum of absolute deviations for the new data. Once again it is the same value as the green circle.

Now move the blue circle to somewhere between 3 and 4 and again find the value that minimizes the sum of absolute differences. This time it is the value of the blue circle.

How do you know which point it will be? The correct placement of the bar will always be at the value of the circle with the middle value. That is, the circle that has two point higher than it and two points lower than it.

Why is this? If the bar is at the circle with the middle value, then moving the bar will bring the bar closer to two points but farther from three points. So, any movement of the bar from the middle value increases the sum of absolute deviations.