An artist arranges 1000 dots evenly around a circle, with each dot being either red or blue. A critic looks at the artwork and counts faults: each time two red dots are adjacent is one fault, and each time two blue dots are exactly two apart (that is, they have exactly one dot in between them) is another.
What is the smallest number of faults the critic could find?
Solution
Placing red then blue dots two at a time round the circle (..RRBBRRBB ...), gives 250 'red' faults and no 'blue' faults.

It remains to prove that every possible configuration contains at least 250 faults.
In every five consecutive dots there must be at least one fault (if any of the middle 3 dots are red, then either that dot has a red neighbour or both its neighbours are blue, both giving a fault; if none of the middle 3 dots are red, then we have 2 blue dots that are 2 apart, giving a fault).
Now divide the 1000 dots into blocks of five consecutive dots, each block overlapping by one dot at each end. There are 250 such blocks, each containing at least one fault. Since each fault consist of at least two dots, and blocks overlap by at most one dot, it is impossible to count the same fault in two blocks. Hence, there must be at least 250 faults.

A slight variation on this argument is to divide the 1000 dots into non-overlapping blocks of four dots and argue that each such block either contains a fault, which we will call an internal fault, or is $R B B R$. In the latter case then the dot that comes directly after the block will be the second dot in a red fault if it is red and will be the second dot in a blue fault if it is blue. Either way we have a fault the first of whose dots is in our block, which we call an external fault. Since each of the 250 blocks contributes either an internal or an external fault, there must be at least 250 faults in any configuration.

Alternative
There are various alternative approaches to showing every configuration has at least 250 faults, such as the following: without loss of generality, we may consider a configuration with the minimum number of faults and no more than two consecutive dots the same colour. Indeed, if there are three or more consecutive blue dots, replace one in the middle with a red dot (this doesn't add any faults, and might remove some faults made of two blue dots that are two apart); if there are three or more consecutive red dots, replace one in the middle with a blue dot (this removes two faults made of two adjacent red dots, adds at most two faults made of blue dots two apart). This step increases the number of blocks of consecutive dots of the same colour, so it must terminate after finitely many steps, at which point all such blocks have length at most 2 .
Now that all blocks of consecutive dots the same colour have length 1 or 2 , the number of faults equals the number of such blocks of red dots (a block of two red dots has a fault in it; a block of one red dot gives a fault of the two blue dots either side). Since there are at least 500 blocks, of alternating colours, there are at least 250 blocks of red dots, so at least 250 faults.