Any number with the sum of the digits equal to $1$ is a power of $10$, so it cannot be a multiple of $7$. Let us try and find a multiple of $7$ such that the sum of its digits will be equal to $2$. This number must have two digits equal to $1$. We check the first few positive integers with this property. The numbers $11$, $101$ and $110$ are not divisible by $7$, but $1001$ is.

The positive integer $10011001\ldots1001$, made by $k$ repetitions of the number $1001$, has the sum of the digits equal to $2k$ and is obviously a multiple of $1001$. Hence, it is also a multiple of $7$. We conclude that for all even positive integers $n$ there exists an integer $m$, which is a multiple of $7$ with the sum of its digits equal to $n$.

It is easy to find a multiple of $7$ with the sum of the digits equal to $3$. The number $21$ has these properties. Any number of the form $2110011001\ldots1001$, which consists of $21$ followed by $k$ repetitions of $1001$, has the sum of the digits equal to $3 + 2k$ and is a multiple of $7$. So, for all odd positive integers $n > 1$ there exists an integer $m$, which is a multiple of $7$, such that the sum of the digits of $m$ is equal to $n$.

We have shown that all positive integers $n$ except $1$ have the required property.