Any number with the sum of the digits equal to $1$ is a power of $10$, so it cannot be a multiple of $13$. Let us try and find a multiple of $13$ such that the sum of its digits will be equal to $2$. This number must have two digits equal to $1$ and the remaining digits must be $0$. We check the first few positive integers with this property. The numbers $11$, $101$ and $110$ are not divisible by $13$, but $1001$ is.

Similarly, let us try and find a positive integer $m$ with the sum of the digits equal to $3$. Let $m = 10^a + 10^b + 10^c$ where $a \ge b \ge c$. The remainder of $10^2$ when divided by $13$ is $9$. The remainder of $10^3$ when divided by $13$ is $12$. For $10^4$ we get $3$, for $10^5$ we get $4$ and for $10^6$ we get $1$. The sum of the remainders given by $10^2$, $10^4$ and $10^6$ is $13$, so $13$ divides $101010$.

Any number of the form $1001\,1001\ldots1001$ with $k$ repetitions of $1001$ is a multiple of $1001$, so it is also a multiple of $13$. The sum of its digits is equal to $2k$. Any number of the form $101010\,1001\ldots1001$ where $101010$ is followed by $k$ repetitions of $1001$ is also a multiple of $13$ and has the sum of the digits equal to $3 + 2k$.

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