Squares

There are $64$ possible $3x3$ grids, and from left to right on the top, their sums are $108$ , $117$ , $126$ , ....

Similarly, each row of $8$ such $3x3$ grids is an arithmetic progression that increases by $9$ each time... The $40$ has had $27$ subtracted, so any $3x3$ grid three to the left of any $3x3$ grid containing the $13$ will suffice. There are $3$ such grids.

By the same argument, for any $3x3$ grid containing the $40$ , a corresponding $3x3$ grid can be found $3$ to the right of it. And there are $6$ such grids.

Finally we have the box containing 13 in the lower right matching the box containing 40 in the top middle.

So in total we have $6+3+1 = \boxed{10}$ ways.