The 35th Japanese Mathematical Olympiad

$72$

Since the sum of the integers in any two adjacent cells must be at most $10$, the only possible integers that can appear in the neighbors of the cell containing $7$ are $1$, $2$, or $3$. Therefore, the cell labeled $7$ cannot be the central cell, and there are exactly $6$ possible cells in which to place the $7$.

By symmetry, the total number of labelings is $6$ times the number of labelings in which the leftmost cell contains $7$. Suppose the leftmost cell contains $7$, and denote this cell by $A$. We classify the other six cells into three groups, $B$, $C$, and $D$, as shown in the left diagram below.



The three cells in group $B$ are all adjacent to $A$, so they must be labeled with $1$, $2$, or $3$ in some order. The remaining three cells in groups $C$ and $D$ must be labeled with $4$, $5$, and $6$. Since the two cells labeled $5$ and $6$ are not adjacent to each other, both of them must lie in group $C$ (which consists of two cells), and the cell in group $D$ must be labeled $4$.

Conversely, any labeling satisfying these assignments clearly meets the condition that each pair of adjacent cells sums to at most $10$. There are $3!$ ways to assign $\{1,2,3\}$ to the three cells in $B$, and $2!$ ways to assign $\{5,6\}$ to the two cells in $C$. Hence the total number of labelings is $6\cdot (3!\cdot 2!)=72$.