Let $\kappa$ be the number of digits of the integer $A$ which is equal to $13$ times the sum of its digits. The least possible $A$ is $10^{\kappa-1}$, while the maximal possible sum of the digits is $9\kappa$. Therefore we need to have:
$$10^{\kappa-1} \leq 13 \cdot 9\kappa = 117\kappa. \qquad (1)$$
For $\kappa \ge 4$, we will prove using induction that: $10^{\kappa-1} > 117\kappa$, that is, relation (1) is not valid. In fact, for $\kappa = 4$ we have: $10^{4-1} = 10^3 > 117 \cdot 4 = 468$. If $10^{\kappa-1} > 117\kappa$, for the arbitrary $\kappa > 4$, then we get:
$$10^{(\kappa+1)-1} = 10^{\kappa} = 10 \cdot 10^{\kappa-1} > 10 \cdot 117\kappa = 117 \cdot 10\kappa > 117 \cdot (\kappa+1).$$
Therefore the number $\kappa$ must be less or equal to $3$.

* We have to reject the case $\kappa = 1$, since $A = \alpha < 13\alpha$, with $0 < \alpha \le 9$.
* Similarly we reject the case with $\kappa = 2$, since $A = 10\alpha + \beta < 13(\alpha + \beta)$.
* Let $\kappa = 3$ and $A = \overline{\alpha\beta\gamma} = 100\alpha + 10\beta + \gamma$, $0 < \alpha \le 9$, $0 \le \beta, \gamma \le 9$.

Then we have:
$$100\alpha + 10\beta + \gamma = 13 \cdot (\alpha + \beta + \gamma), \quad 0 < \alpha \le 9, 0 \le \beta, \gamma \le 9$$
$$\Leftrightarrow 87\alpha = 3\beta + 12\gamma, \quad 0 < \alpha \le 9, 0 \le \beta, \gamma \le 9$$
$$\Leftrightarrow 29\alpha = \beta + 4\gamma, \quad 0 < \alpha \le 9, 0 \le \beta, \gamma \le 9$$
Since $0 \le \beta + 4\gamma \le 45 \Rightarrow 29\alpha \le 45 \Rightarrow \alpha \le 1$ (since $\alpha \ne 0$), and hence:
$$\beta + 4\gamma = 29 \Rightarrow \beta = 29 - 4\gamma \ge 0 \Rightarrow 0 \le \beta = 29 - 4\gamma \le 9$$
$$\Rightarrow -29 \le -4\gamma \le -20 \Rightarrow 5 \le \gamma \le \frac{29}{4} \Rightarrow \gamma \in \{5, 6, 7\}$$
* For $\gamma = 5 \Rightarrow \beta = 9$ and $A = 195$.
* For $\gamma = 6 \Rightarrow \beta = 5$ and $A = 156$.
* For $\gamma = 7 \Rightarrow \beta = 1$ and $A = 117$.

Therefore, the positive integers which are equal to $13$ times the sum of their digits are $117$, $156$, and $195$.