Question: Can you place seven (point-sized) pawns on a $7\times7$ checkerboard, so that

* every pawn is placed precisely in the middle of one of the little checkerboard squares, and
* all distances between pairs are different?

Answer: 
Not really an answer but I'd like to add my thought process to this

> 
>  First I checked: How many possible distances are there really on a 7x7 checkerboard? To do this you can draw lines from (1,1) to (1,7) till (7,7), from (1,1) to (1,6) till (6,6) and so on. This mean $7+6+5+4+3+2 = 27$, However because 3,4,5 is a pythagorean triple it is one less because (1,1) to (4,5) = (1,1) to (6,1). There are 21 connections between 7 points. My intuition therefore says it is very unlikely that there are 21 connections that are all different because there are only 26 possibilities to begin with.  
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> This does tell us that if there is a solution that at least 3 points are in a square along the border. Why? There are 7 distances that go from border to border and only a maximum of 5 are not used of these. The two distances could share a point, therefore at least 3 points are on the border squares
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I would say:

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>  No this isn't possible. You should only have 26 different options of distance between pawns and the total number of relationships to take into account is 5040. Still thinking on it, but if I'm right feel free to tell me. 
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This is unless:

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>  The board itself isn't square ;)
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