XXVII Olimpiada Matemática Rioplatense

It is easy to see that $A-B\ne 0$. Indeed, we have $A-B=0$ if and only if the $27$ sums are all equal; however, if $x,y,z$ are three consecutive numbers on the circumference, the sums $x+y$ and $y+z$ are different, since $x\ne z$.

We will now prove that it is not possible that $A-B=1$. If this is the case, when considering three consecutive numbers $x,y,z$ on the circumference, the sums $x+y$ and $y+z$ differ by $1$, since they are not equal, and so, $x$ and $z$ differ by $1$. Now look at the place where the number $27$ is located; assume the previous numbers are $a,b$ and the following are $c,d$.



By the previous arguments, $27$ and $d$ differ by $1$, and the same happens with $27$ and $a$. But the only written number that differs by $1$ with $27$ is $26$; then, both $a$ and $d$ would be equal to $26$, a contradiction. This proves that we cannot achieve $A-B=1$.

Finally, we show an example where $A-B=2$.

First, Lucía writes number $1$ and then, she alternates odd and even numbers, writing odd numbers in decreasing order and even numbers in increasing order.

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The first sum is $1+27=28$ and, the following ones are alternately $29$ and $27$. Therefore, $A-B=29-27=2$, as stated.