Royal attack

There is one way to arrange four kings (one in the middle of each edge),

And wiith three kings you can fill any three squares of any of the 2x2 sub-grids, and for each of the 4 sub-grids there are 4 ways to choose the vacant square, for a total of 16 ways.

This gives a total of $1+16 = \boxed{17}$ ways.

Place a king in one of the corners. Then, 2 of the three squares adjacent must also contain kings. Since all three squares are in range of each other, these additional kings will be already satisfied. There therefore cannot be any other kings on the board, as they would be adjacent to one of these three. There are 4 corners, and three ways of picking the other two, so there are 12 cases with a king in a corner.

Any remaining cases don’t use the corners, so we are now searching a cross with five spaces. If we place one in the center, all four remaining spaces touch it, so only two more kings can be placed. Since they must also touch each other, to satisfy their own requirements, there are 4 of these cases.

Finally, if we say there are no kings in the corners or center, we have one more case where all four remaining spaces have kings. So, the total is 12 + 4 + 1 = $\boxed{17}$