Tables of 1 to N with Equal Columns

For any composite integer N=mn we can arrange the integers 1 to N in 
a rectangular array with m rows and n columns such that the sum of
each individual column equals N(N+1)/(2n).  Naturally if there are
an even number of rows this is trivial, since we can simply list the 
numbers sequentially left-to-right on odd-numbered rows and right-
to-left on even-numbered rows, such as the 2x3 table

                       1 2 3
                       6 5 4

However, with an odd number of rows (greater than 1) we need to use 
a slightly more subtle arrangement.  We can also assign the numbers
1 through n (in order) to the first row, i.e., we set 

                 A(1,k) = k            k = 1,2,..,n

The second row is assigned the numbers n+1 through 2n, placed
consecutively beginning in the column just to the right of center,
proceding to the right-most column, and then continuing from the
left-most column to the center.  Thus we set

                   /  k + (n-1)/2 + n    for k=1 through (n+1)/2
       A(2,k)  =  (
                   \  k + (n-1)/2        for k = (n+3)/2 through n

The third row contains the numbers 2n+1 through 3n.  The number 2n+1
is placed in the middle column, and 2n+2 is placed in the right-most
column.  Then 2n+3 is placed just to the left of 2n+1, and then 2n+4
is placed just to the left of 2n+2, and so on, alternating until the
row is filled.  These assignments can be expressed as

                   /  3n + 2 - 2k     for k = 1 through (n+1)/2
       A(3,k)  =  (
                   \  4n + 2 - 2k     for k = (n+3)/2 through n

Notice that each column from k=1 to (n+1)/2 has the following sum
over the first three rows

 A(1,k) + A(2,k) + A(3,k)  =  [k]  +  [n+(n-1)/2+k]  +  [3n + 2 - 2k]

                           = (9n+3)/2

whereas for the columns from k=(n+3)/2 to n the sums have the form

 A(1,k) + A(2,k) + A(3,k)  =  [k]  +  [(n-1)/2 + k]  +  [4n + 2 - 2k]

                           =  (9n+3)/2

Thus every column sums to the same value, (9n+3)/2, over these first
three rows.  This leaves an even number of rows to be filled with the
numbers 3n+1 to N, so these can be assigned consecutively in the
usual pattern for pairs of rows.  To illustrate this pattern, here
are two tables for N=35:

            n=5,m=7                  m=7,n=5

          1  2  3  4  5        1  2  3  4  5  6  7
          8  9 10  6  7       11 12 13 14  8  9 10
         15 13 11 14 12       21 19 17 15 20 18 16
         16 17 18 19 20       22 23 24 25 26 27 28
         25 24 23 22 21       35 34 33 32 31 30 29
         26 27 28 29 30
         35 34 33 32 31

This raises some interesting questions.  For example, is it always
possible to construct a "primitive" m x n equal-column array, meaning 
that no subset of the rows form an equal-column array.  We've seen
that there are primitive EC arrays for 2 and 3 rows, but what about
arrays with more than 3 rows?