Around

We arrange the numbers in any way and arrive at the conclusion that some sum of six consecutive numbers is bigger than 300:

Let the sum of any six consecutive numbers be $S$ and let $S'$ be the sum of the other five numbers plus one number $x$ from the previous six that is adjacent to the last five. Then, by construction, $S+S'= 550 + x$ where the first term of the right hand side comes from the sum of all the numbers from $45$ to $55$ . Because the numbers bigger than 50 have to appear in the circles, there are selections of $S$ and $S'$ that have these numbers as their intersection $x$ . In those cases, $S+S'>600$ . Therefore, one of the sums $S,S'$ is bigger than 300

The sum of the 11 given numbers is 550. Suppose they are arranged as required by the question.

If we let S(i), with i being 1..11, be the sum of the six numbers starting at 44+i and moving clockwise then:

The sum of these eleven sums is 6*550 = 3300 because each number has been counted 6 times

So the mean of the 11 sums is 300. But none are above 300, so none are below. So all eleven sums are 300.

In particular: the set of six circles starting at 50 will sum to 300. So 5 consecutive circles sum to 250. They can only be completed to 300 on one side.