Blocksum-Simple Permutations

Let the permutation be $a_1, a_2, a_3, \ldots, a_{16}$ . We group these as below: $\begin{aligned} a_1, a_5, a_9, a_{13}\\ a_2, a_6, a_{10}, a_{14}\\ a_3, a_7, a_{11}, a_{15}\\ a_4, a_8, a_{12}, a_{16} \end{aligned}$ Note that because of the above property, each of the rows consist of consecutive numbers, and two adjacent rows are consecutive in opposite directions.

Thus, there are $2 \cdot 4! = \boxed{48}$ ways of arranging the numbers such that we get a blocksum-simple permutation.

An example of a satisfying permutation is $1,8,9,16,2,7,10,15,3,6,11,14,4,5,12,13$ , as seen below:

$\begin{array}{c}1 & 2 & 3 & 4\\ 8 & 7 & 6 & 5\\ 9 & 10 & 11 & 12\\ 16 & 15 & 14 & 13 \end{array}$