Python code for serving k customers with bounded time window
Time complexity is unnecessarily large.
First, it is linear in the timespan of the arrivals. A list [[1,2], [100,101]] would require 100 iterations; a list [1,2], [1000,1001]] would require 1000 iterations, etc. Don't bruteforce every minute.
Second, del intervals[0] has a linear time complexity in number of intervals, which results in an overall quadratic time complexity. A cheap way out is to sort in a descending order, and work right to left. This way you would remove intervals from the tail of the list, which is O(1).

DRY. A presence of two semantically identical breaks (the loop is broken when it runs out of the intervals) suggests refactoring. Do not test for if not intervals explicitly. Consider
  try:
      while i &lt; end:
          for j in range(k):
              if intervals[0][0] &lt;=i &lt;= intervals[0][1]:
                  del intervals[0]
          i += 1
  except:
      pass

Of course, never except blindly. Catch only exceptions you expect and know how to handle, in this case except IndexError:.