Write each integer in base N, that is each x can be represented as (x1, x2) with x = 1 + x1 + x2*N. Now you can sort it twice with counting sort, once on x1 and once on x2, resulting in the sorted array.
EDIT: As others mentioned below, sorting on each ‘digit’ separately like this is is called a radix sort. Sorting on each digit with counting sort takes O(N) time and O(N) space (in this particular case). Since we repeat it exactly twice, this gives a total running time of O(N).
Yes, you can, using radix sort with N buckets and two passes. Basically, you treat the numbers as having 2 digits in base N.