Hi, I am afraid I am reinventing the wheel here, but I had fun writing it, so here it goes. Comments:
1. It's just a binary search, as suggested by everyone else!
2. The performance should be quite good, but I did not test it against big arrays (and I would obviously need to write something to test against...)
3. In case two numbers in the array are equally close, the bigger number will be chosen as the winner.
Update: Of course I have to assume the target array is sorted.

use strict;
use warnings;

my @array = (1,4,5,6,7,9,10,23,34,44,55,56,57,59,70,80,90,100);

print "\nenter number\n";
chomp (my $find = <STDIN>);

my $nearest = @{nearest(\@array)}[0];
print "nearest to $find in array is: $nearest\n";

sub nearest {
  my ($a) = @_;  
  my $size = @$a;
  return $a if $size == 1;
  my $mid = int(($size-1) / 2);
  my $test = @$a[$mid];  
  return $test <= $find ?
    (abs($test-$find)<abs(@$a[$mid+1]-$find) ? [$test] : 
       $find <= @$a[$mid+1] ? [@$a[$mid+1]] : nearest([@$a[$mid+1 .. $
+size-1]]))
    :
    (abs($test-$find)<abs(@$a[$mid-1]-$find) ? [$test] : 
       $find >= @$a[$mid-1] ? [@$a[$mid-1]] : nearest([@$a[0 .. $mid]]
+));
}
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