Delete lines with specific pattern name in the first column awk - awk

Question

I want to delete all lines in which MGD is not between 676 and 900.
I wrote a script
#!/bin/bash
for index in {1..100} # I do this script on 100 files, that is why I use for loop
do
awk 'BEGIN { FS = "MGD" };
{if ($2 >= 676 && $2 <= 900) print}' eq2_15_333_$index.ndx | tee eq3_15_333_$index.ndx
done
Input example
MGD816 SOL77
MGD71 SOL117
MGD7 SOL13194
MGD18 SOL235
MGD740 SOL340
MGD697 SOL396
MGD70 SOL9910
Expected output
MGD816 SOL77
MGD740 SOL340
MGD697 SOL396
I don't know what my script do something wrong, because I still have something which has MGD7 or MGD71, but MGD18 I haven't in my output.
Edit
I tested this script and it works perfectly
awk '/^MGD/{val=substr($1,4);if(val+0 >= 676 && val+0 <= 900){print}}' new.txt | tee new2.txt
and I have output
MGD816 SOL77
MGD740 SOL340
MGD697 SOL396

Answer 1

Based on your shown samples try following once. This is completely based on your shown attempt, since no samples were provided so its not tested, should work though.
#!/bin/bash
for index in {1..100}
do
awk '/^MGD/{val=substr($1,4);if(val+0 >= 676 && val+0 <= 900){print}}' eq2_15_333_$index.ndx | tee eq3_15_333_$index.ndx
done

Answer 2

I want to explain why your original i.e.
awk 'BEGIN { FS = "MGD" }; {if ($2 >= 676 && $2 <= 900) print}'
did not work as excepted, you set "MGD" as FS thus AWK splitted only at MGD - if you do awk 'BEGIN{FS="MGD"}{print $2}' file.txt and content of file.txt is
MGD816 SOL77
MGD71 SOL117
MGD7 SOL13194
MGD18 SOL235
MGD740 SOL340
MGD697 SOL396
MGD70 SOL9910
output is
71 SOL117
7 SOL13194
18 SOL235
740 SOL340
697 SOL396
70 SOL9910
If you want $2 to be first number you should specify FS which is "MGD" or spaces i.e.
awk 'BEGIN{FS="MGD|[[:space:]]+"}...