sum of its digits?

There can't be any one-digit solution, because every one-digit number is equal to the sum of its digits.

Let's explore the case for two-digits numbers.

Let the $a$ and $b$ be the digits of the number, which can be expressed as $10a+b$ .

$10a+b=7(a+b)$ , as stated in the question.

$10a+b=7a+7b$ , by expanding the product.

$10a-7a=7b-b$ , by rearranging the terms.

$3a=6b$

$a=2b$

An integer solution is needed and the smallest is $a=2$ and $b=1$ .

The solution is $21$ , and the conditions are verified.