Use sed to add character in n'th position of matching string

Another approach:

$ sed 's/"\([^"]*blah[^"]*"\)/"N\1/g' test.txt 
"N1blah8","na","N8blah4"  
"N2blah5","na","N10blah4"  
"N5blah5","na","1blah234

The regex is looking for a ", then 0 or more non-" characters followed by a blah, and then then 0 or more non-" again. Because of the parentheses, this is captured and can later be referred to as \1. Therefore, the command will substitute the matched pattern with itself (\1) but with a "N appended. That's why the first " is outside the parentheses. The /g modifier at the end makes it subsstitute all matching string in each line.

If your sed version supports it, you can simplify it to:

sed -E 's/"([^"]*blah[^"]*")/"N\1/g'

To get sed to repeat the substitution for all matches on a line, add the g flag after the last /:

$ sed 's/[^"]*blah/N&/g' test.txt

The g flag will cause the substitution to be made "for all non-overlapping matches of the regular expression, not just the first one" (quote from the manual on my system).

I've also changed the regular expression a tiny bit so that all the characters that are not " before blah are matched. This way the substitution will insert the N just before the first digit, just after the ".

If you wanted to insert after the nth character (where n is an arbitrary value) then you should avoid sed. There are better tools for this job (like awk or perl, python etc) e.g. with awk you can use a variable n to insert STRING after the nth character in each field that matches PATTERN:

awk -vn=2 'BEGIN{FS=OFS=","}{for (i=1; i<=NF; i++){ if ($i ~ /PATTERN/){
h=substr($i,1,n);t=substr($i,n+1,length($i));$i=h "STRING" t}}};1' infile

This assumes your file is a simple csv file (no commas embedded in your fields).

CSV parsing and processing may be real complex. Here is a perl oneliner which put the N in front of column one and three, independently of the content:

Step 1, strip the double quotes:

@F = map { /"(.*)"/ } @F;

Step 2, inserting the N on column one and three

$F[0,2] =~ s/^/N/;

Step 3, join the array as CSV

printf qq("%s"\n), join q(","), @F

run perl:

perl -F, -ane '@F = map { /"(.*)"/ } @F; $F[0,2] =~ s/^/N/; printf qq("%s"\n), join q(","), @F' csvfile

Edit: new approach to avoid the security risk of eval expression (thank you terdon).