grep has no relevance here.

$|-- and --$| toggle the current value.
$|++ and ++$| set the value to 1.
$|-- and $|++ return the old value.
--$| and ++$| return the new value.

Normally, $| is set to 0.

For example,

print('$|--: '); $|=0; for (1..10) { print($|--, ' '); } print("\n");
print('--$|: '); $|=0; for (1..10) { print(--$|, ' '); } print("\n");
print('$|++: '); $|=0; for (1..10) { print($|++, ' '); } print("\n");
print('++$|: '); $|=0; for (1..10) { print(++$|, ' '); } print("\n");
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outputs

$|--: 0 1 0 1 0 1 0 1 0 1
--$|: 1 0 1 0 1 0 1 0 1 0
$|++: 0 1 1 1 1 1 1 1 1 1
++$|: 1 1 1 1 1 1 1 1 1 1
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Update: $toggle^=1 works just as well as --$|, but it's less obfuscated and has no side-effects:

$,=", ";
$\="\n";
@list = qw( a b c d );

$| = 0;
print grep --$|,  @list;  # a, c

$t = 0;
print grep $t^=1, @list;  # a, c
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Update: If you need every Nth term, try the following:

$,=", ";
$\="\n";
@list = qw( a b c d e f );

$n = 3;
$t = $n-1;
print grep !($t=($t+1)%$n), @list;  # a, d
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Thie following code is instructive:

use strict;
use warnings;

$| = 0;
print $|++ . " " for (1..4);

$| = 0;
print "\n";
print $|-- . " " for (1..4);
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Prints:

0 1 1 1 
0 1 0 1
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Why does that happen? Well, from the documentation snippet referred to in Perl Idioms Explained - $|++, you see that the only legal values for $| are 0 and 1. When you increment $| to 2 it is set back to 1, so the 1 'sticks'. However, when you decrement $| to -1 the -1 is changed to 1. Next time it decrements from 1 to 0 - back where you started. A cheap toggle, so long as you are not interested in the effects on buffering.

You might also consider:

use strict;
use warnings;

my $x = 1;
print $x ^= 1 . " " for (1..4);
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which prints:

0101
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If you want $| to be 1, assign 1 to it. Pre- or post indcrement is unnecessarily "clever" and could bite you if $| doesn't happen to have the value you assume.

Someone seems intent on repeating this particular piece of misinformation, so perhaps it's worth repeating (just this once) that it's incorrect.

I do agree that $|++ is unnecessarily clever, since it might confuse an ignorant maintenenance programmer, and that $| = 1 is clearer. But it's simply wrong to say that it could “bite you if $| doesn't happen to have the value you assume”. It couldn't, for reasons that will be obvious to anyone who has understood this thread.