, 2018

Solution:
Let the four numbers be $a, b, c, d$ around the square. Assume without loss of generality that $a$ is the largest number, so that $a > b$ and $a > d$. Note that $c$ cannot be simultaneously smaller than one of $b, d$ and larger than the other because, e.g. if $b > c > d$, then $a > b > c > d$ and $a \geq d + 12$. Hence $c$ is either smaller than $b$ and $d$ or larger than $b$ and $d$.

Case 1: $c$ is smaller than $b$ and $d$. Then we have $a - c \geq 8$, but when $a - c = 8$, we have $b = c + 4 = d$, so we need $a - c = 9$, giving the only set $\{1, 5, 6, 10\}$.

Case 2: $c$ is larger than $b$ and $d$. Since $a > c$ and $b, d$ are both at most $c - 4$, the range of possible values for $c$ is $\{6, 7, 8, 9\}$. When $c = 9, 8, 7, 6$, there are $1, 2, 3, 4$ choices for $a$ respectively and $\binom{5}{2}, \binom{4}{2}, \binom{3}{2}, \binom{2}{2}$ for $b$ and $d$ respectively (remember that order of $b$ and $d$ does not matter). So there are $1 \cdot 10 + 2 \cdot 6 + 3 \cdot 3 + 4 \cdot 1 = 35$ sets in this case.

Therefore we have $1 + 35 = 36$ possible sets in total.