Harvard-MIT Mathematics Tournament

Solution:

First note that all the integers must be at least $2$, because the greatest common divisor of any two positive integers is at least $1$. Let $n$ be the largest integer in the circle. The greatest common divisor of its two neighbors is $n-1$. Therefore, each of the two neighbors is at least $n-1$ but at most $n$, so since $n-1 \nmid n$ for $n-1 \geq 2$, they must both be equal to $n-1$.

Let $m$ be one of the numbers on the other side of $n-1$ from $n$. Then $\operatorname{gcd}(n, m) = n-2$. Since $n-2 \geq 0$, $n-2 \mid n$ only for $n=3$ or $4$.

If $n=3$, each number must be $2$ or $3$, and it is easy to check that there is no solution.

If $n=4$, then it is again not hard to find that there is a unique solution up to rotation, namely $4\ 3\ 2\ 2\ 3\ 4\ 3\ 2\ 2\ 3$. The only possible sum is therefore $28$.