#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

my ($START, $END, $SIZE) = (1, 1524, 150);

my @range;
while (my $line = <DATA>) {
    my ($pos, $from, $to) = split / |\.\./, $line;
    if ($START == $from && $END == $to) {
        @range[$pos .. $SIZE + $pos - 1] = (1) x $SIZE;
    }
}

say scalar grep $_, @range;

__DATA__
46 1..1524
832 1..1524
1008 1..1524
1407 1..1524
2360 2052..3260
2967 2052..3260
403 1..1524
800 1..1524
2986 2052..3260
3170 2052..3260
[download]

UPDATE Moreover, it seems your description ("Making the array 1 if it contains the position+150 else the array is set to 0") should produce a different output, as you only want to count the part of the 1407 that overlaps 1..1524. The following code does that, and show an alternative approach, without using the array - it uses a hash to remember the positions where the state of the element would change.

#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
use Syntax::Construct qw{ // };


my ($START, $END, $SIZE) = (1, 1524, 150);

my %borders;

while (my $line = <DATA>) {
    my ($pos, $from, $to) = split / |\.\./, $line;
    if ($START == $from && $END == $to) {
        $borders{$pos}++;
        $borders{ $pos + $SIZE + 1 }--;
    }
}

my ($sum, $step) = (0, 0);
for my $i ($START .. $END) {
    $step += $borders{$i} // 0;
    $sum++ if $step;
}

say $sum;

__DATA__
46 1..1524
832 1..1524
1008 1..1524
1407 1..1524
2360 2052..3260
2967 2052..3260
403 1..1524
800 1..1524
2986 2052..3260
3170 2052..3260
[download]

($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
[download]