Prove that among any 18 consecutive 3-digit numbers there must be at least one which is divisible by the sum of its digits.

(In reply to

solution by Praneeth)

Absolutely true, but case 1 is included in case 2.

Erasing the first four rows and the words "case 2" does not affect the validity of your proof.

Your solution boils down to a shorter text:

In **a row** of 18 consecutive **numbers** two are divisible by 9:

one is an even multiple and the other is odd.

Both 2 and 9 divide the even one.

q.e.d.