if statement query user id in etc/passwd?

To do that in a bash loop, I suggest using read and IFS like:

#!/usr/bin/env bash
while IFS=':' read -r user passwd uid gid comment home shell;  do
    if [ "$uid" -gt 1000 ] ; then
        echo GT $user
    else
        echo LT $user
    fi
done < /etc/passwd

Instead of reading /etc/passwd directly, you should use getent passwd, that will also work if some of your users are saved in something like LDAP or such. awk should be well-suited for the colon-separated output format.

This would print the usernames of all users with UID > 1000:

getent passwd | awk -F: '$3 > 1000 {print $1}'

And this would just print found if at least one such is found:

getent passwd | awk -F: '$3 > 1000 {print "found"; exit}'

Try this:

if grep -E '^[^:]*:[^:]*:[^:]{4}' /etc/passwd | grep -Evq '^[^:]*:[^:]*:1000:'

The first grep searches passwd for lines with a uid of four or more digits. The second grep filters out the line with uid 1000. The exit status will be 0 if any lines remain, 1 if not.