Since your dates appear to be in a nice comparable format, you can just do ordinary string comparisons on them to find out which is oldest. So ... something like this should work fine:

#!/usr/bin/perl

use strict; use warnings;

my (%oldest_date, %oldest_entry);
while (<DATA>)
{
    my ($item,$group,$date) = split;
    if (!exists $oldest_date{$group} || $date lt $oldest_date{$group})
+ {
        $oldest_date{$group} = $date;
        $oldest_entry{$group} = $_;
    }
}

for my $g  (keys %oldest_entry) {
    print $oldest_entry{$g};
}


__DATA__
34     gr1    2003-03-02
12     gr1    1990-03-14
39     gr3    2002-04-11
66     gr4    2006-03-16
32     gr3    1998-02-13
90     gr1    2004-06-15
55     gr4    1999-06-15
[download]

use List::Util qw( maxstr);

my %tb;
while ( <DATA> ) {
    my ( undef, $group, $entry_date) = split;
    $tb{ $group}->{ $entry_date} = $_;
}

print $_->{ maxstr keys %$_} for values %tb;

__DATA__
34     gr1    2003-03-02
12     gr1    1990-03-14
39     gr3    2002-04-11
66     gr4    2006-03-16
32     gr3    1998-02-13
90     gr1    2004-06-15
55     gr4    1999-06-15
[download]

Anno

#!/usr/bin/perl -w
use strict;

my $entries;

while ( my $line = <DATA> ){
        $line =~ /\d?\W*(gr\d)\W*(\d*-\d\d-\d\d)/;
        next if ( !$2 );

        my $group = $1;
        my $date = $2;
        $date =~ s/-//g;

        if ( ! defined( $entries->{$group})   || 
                ( $entries->{$group}->{date} < $date )  ){
                $entries->{$group}->{date} = $date;
                $entries->{$group}->{entry} = $line;
        }

}

foreach (keys( %{$entries} )){
        print "entry: $entries->{$_}->{entry}";
}




__DATA__
item  group  entry_date
34     gr1    2003-03-02
12     gr1    1990-03-14
39     gr3    2002-04-11
66     gr4    2006-03-16
32     gr3    1998-02-13
90     gr1    2004-06-15
55     gr4    1999-06-15
etc ...
[download]

#!/usr/bin/perl -w


open F, ">testdata";



for ( 0..1000000 ){
        print F "$_  gr".int(rand(10))."  ". (1990+int(rand(25))) . '-
+0'. (int(rand(10))) . '-' . (10 + int(rand(20)) )."\n";
}
close F;
[download]

my code:

time ./latestentries.pl
entry: 15970  gr5  2014-09-29
entry: 79485  gr8  2014-09-29
entry: 135788  gr7  2014-09-29
entry: 221  gr2  2014-09-29
entry: 18669  gr9  2014-09-29
entry: 46760  gr1  2014-09-29
entry: 4960  gr3  2014-09-29
entry: 9486  gr0  2014-09-29
entry: 19710  gr4  2014-09-29
entry: 56757  gr6  2014-09-29

real    0m8.689s
user    0m8.617s
sys     0m0.060s

-------------------
anno's code:

micha@laptop ~/prog/perl/test $ time perl test-anno.pl
962757, gr0, 2014-09-29
964472, gr1, 2014-09-29
984704, gr2, 2014-09-29
980128, gr3, 2014-09-29
985851, gr4, 2014-09-29
931318, gr5, 2014-09-29
976880, gr6, 2014-09-29
988367, gr7, 2014-09-29
992654, gr8, 2014-09-29
962175, gr9, 2014-09-29

real    0m4.556s
user    0m4.424s
sys     0m0.036s

-------------------
and duff's entry:

micha@laptop ~/prog/perl/test $ time perl test-duff.pl
100154  gr5  1990-00-10
5654  gr8  1990-00-10
2318  gr7  1990-00-10
9789  gr2  1990-00-10
19151  gr9  1990-00-10
91314  gr1  1990-00-10
124846  gr3  1990-00-10
14858  gr0  1990-00-10
175946  gr4  1990-00-10
95691  gr6  1990-00-10

real    0m3.497s
user    0m3.452s
sys     0m0.036s
[download]

The reading should be straight forward, you can use split to access the individual fields.

Since you only want to write one item per group, I'd suggest you use a hash with the group as the keys, and every time you read a line you compare if the read date is older than the current date in the hash. If yes, you replace it.

What does happen when you get two identical dates?

As you don't say anything about the context, supposing unix-like, I thought I could mention various one-liners to get a feeling of your data:

% steph@apexPDell2 (/home/stephan/t) %
% cat data.txt # I added the last line                                
+                                       

item  group  entry_date
34     gr1    2003-03-02
12     gr1    1990-03-14
39     gr3    2002-04-11
66     gr4    2006-03-16
32     gr3    1998-02-13
90     gr1    2004-06-15
55     gr4    1999-06-15
10     gr1    2003-03-02

% steph@apexPDell2 (/home/stephan/t) %
% LC_ALL=C sort -k 3 data.txt | perl -lna -e 'print if $F[1] eq q{gr1}
+ and $F[0] == 34'
34     gr1    2003-03-02

% steph@apexPDell2 (/home/stephan/t) %
% sort -k 3 data.txt | grep gr1 | sort -n | head -n1
10     gr1    2003-03-02
[download]

The last one reads as sort on the date, select group gr1, select on the first numerically and keep tghe first line. In this particular case it is faster to grep first.