The 1st column represents a 4 position date (e.g. 10/02). I want print only the lines with succeeding dates (e.g. 10/01, 10/02, 10/15, 10/16, 10/20, 10/21, 10/29 & 10/30).

File named test

Test:
1001 1715 3bc4r
1002 1927 3aclr
1002 2127 01iqq
1002 2227 3j1ai
1004 2021 3x1mq
1006 1947 yj1vd
1011 0007 s7tem
1015 1448 05p2l
1016 1248 3j8nk
1020 2137 0aakg
1021 2138 0aakg
1026 2045 xuakr
1029 2110 3alzt
1029 2310 0aakg
1030 0010 3jf24

Desired Output:

1001 1715 3bc4r
1002 1927 3aclr
1002 2127 01iqq
1002 2227 3j1ai

1015 1448 05p2l
1016 1248 3j8nk

1020 2137 0aakg
1021 2138 0aakg

1029 2110 3alzt
1029 2310 0aakg
1030 0010 3jf24


Any help would be appreciated.

You can do this with awk. Here is the plan (needs some fine-tuning):

Store a line in an array.
Keep adding lines to the array as long as the date is the same as the previous line or one day later.
As soon as the input line has a date that is two or more days later, dump out the array and reset it.

One difficulty is that you need a function that checks if a date follows another. How do you decide that for the last day of February?

a problem with this is that it checks the whole line for the number.

Thanks! It has been 15 year since I had access to a command line, so I am pretty rusty. I'll explore the awk array until something else materializes.