ok, I am just a beginner with AppleScript, and I know this is NOT the fastest or most efficient method, but it seems to work.
but as I teach new students in VBA (that I do know very well), it aint wrong if it works... it just might not be the most efficient, but it isn't wrong.
let me know if this works for your setup, I would like it if it did. :)
Jason
tell application "Numbers"
-- the first three tells here get to the selected range, then use the first cell of that range
-- we are assuming you are using a single cell to delete here, like the OP
tell front document's active sheet
tell (first table whose selection range's class is range)
tell selection range
-- get the row and column numbers of that first cell in the selection
set DelRow to the address of the row of the first cell
set DelCol to the address of the column of the first cell
end tell
-- IF your selection is NOT the last row of the table, continue
if DelRow < address of the row of the last cell of column DelCol then
-- Remember the origianl cells location for later
set firstcell to the name of cell DelCol of row (DelRow)
-- Get the number of the lastrow of the column
set lastrow to the address of the row of the last cell of column DelCol
-- this is the part that might take a little while
-- walk from the starting cell, and set its value to the value of the
-- cell beneath it.
repeat with i from DelRow to lastrow - 1
set value of cell DelCol of row i to value of cell DelCol of row (i + 1)
end repeat
-- now set the value of the last cell in the column to nothing/MT
set value of cell DelCol of row lastrow to ""
-- I felt is useful to reselct the original cell again at the end
set selection range to range firstcell
else
-- only show this message if there is nothing below to bring up
display alert "Last cell of column Selected"
end if
end tell -- end tell of table
end tell -- end tell of document1 front sheet
end tell -- end tell numbers