How can I extract a predetermined range of lines from a text file on Unix?

Quite simple using head/tail:

head -16482 in.sql | tail -258 > out.sql

using sed:

sed -n '16224,16482p' in.sql > out.sql

using awk:

awk 'NR>=16224&&NR<=16482' in.sql > out.sql

sed -n '16224,16482p;16483q' filename > newfile

From the sed manual:

p - Print out the pattern space (to the standard output). This command is usually only used in conjunction with the -n command-line option.

n - If auto-print is not disabled, print the pattern space, then, regardless, replace the pattern space with the next line of input. If there is no more input then sed exits without processing any more commands.

q - Exit sed without processing any more commands or input. Note that the current pattern space is printed if auto-print is not disabled with the -n option.

and

Addresses in a sed script can be in any of the following forms:

number Specifying a line number will match only that line in the input.

An address range can be specified by specifying two addresses separated by a comma (,). An address range matches lines starting from where the first address matches, and continues until the second address matches (inclusively).

sed -n '16224,16482 p' orig-data-file > new-file

Where 16224,16482 are the start line number and end line number, inclusive. This is 1-indexed. -n suppresses echoing the input as output, which you clearly don't want; the numbers indicate the range of lines to make the following command operate on; the command p prints out the relevant lines.

You could use 'vi' and then the following command:

:16224,16482w!/tmp/some-file

Alternatively:

cat file | head -n 16482 | tail -n 258

EDIT:- Just to add explanation, you use head -n 16482 to display first 16482 lines then use tail -n 258 to get last 258 lines out of the first output.