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.