Just to be really clear, this has nothing to do with enumerate and everything to do with list comprehension syntax.

This list comprehension returns a list of tuples:

[(i,j) for i in range(3) for j in 'abc']

this a list of dicts:

[{i:j} for i in range(3) for j in 'abc']

a list of lists:

[[i,j] for i in range(3) for j in 'abc']

a syntax error:

[i,j for i in range(3) for j in 'abc']

Which is inconsistent (IMHO) and confusing with dictionary comprehensions syntax:

>>> {i:j for i,j in enumerate('abcdef')}
{0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e', 5: 'f'}

And a set of tuples:

>>> {(i,j) for i,j in enumerate('abcdef')}
set([(0, 'a'), (4, 'e'), (1, 'b'), (2, 'c'), (5, 'f'), (3, 'd')])

As Óscar López stated, you can just pass the enumerate tuple directly:

>>> [t for t in enumerate('abcdef') ] 
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f')]

Just tell enumerate to start with the start index.

saved_index=[]
for s,e in zip(start, end):
    for i,l in enumerate(filelist[s:e], s):
        if "blah" in l:
            saved_index.append(i)

For your first question: the index starts at 0, as is generally the case in Python. (Of course, this would have been very easy to try for yourself and see).

>>> x = ['a', 'b', 'c']
>>> for i, word in enumerate(x):
    print i, word


0 a
1 b
2 c

For your second question: a much better way to handle printing every 30th line is to use the mod operator %:

if i+1 % 30 == 0:
    print i
    # ...

This is slightly better than testing for membership in range since you don't have to worry about edge effects (i.e. since range gives a half-open interval that is open on the right end).