3. Find all positive integers that are 13 times the sum of their digits.
3. Let $k$ be the number of digits of the integer $A$. Then $A$ is equal to 13 times the sum of its digits. Since the minimum digit sum of $A$ is $10^{k-1}$, and the maximum digit sum is $9k$, it is necessary to prove:
$$
10^{k-1} \leqslant 13 \times 9 k = 117 k.
$$

We will prove by mathematical induction that for $k \geqslant 4$,
$$
10^{k-1} > 117 k.
$$

When $k=4$, $10^{4-1} = 10^3 > 117 \times 4 = 468$.
For any $k > 4$, if $10^{k-1} > 117 k$, then
$$
\begin{array}{l}
10^{(k+1)-1} = 10^k = 10 \times 10^{k-1} \\
> 10 \times 117 k = 117(k+1).
\end{array}
$$

This contradicts the inequality (1).
Therefore, the positive integer $k \leqslant 3$.
Since $A = a < 13a (0 < a \leqslant 9)$, $k \neq 1$.
Similarly, $k \neq 2$.
$$
\begin{array}{l}
\text{Let } k=3, A=\overline{abc} = 100a + 10b + c, \text{ where,} \\
0 < a \leqslant 9, 0 \leqslant b, c \leqslant 9.
\end{array}
$$
$$
\begin{array}{l}
\text{Then } 100a + 10b + c = 13(a + b + c) \\
\Leftrightarrow 87a = 3b + 12c \\
\Leftrightarrow 29a = b + 4c.
\end{array}
$$

Since $0 \leqslant b + 4c \leqslant 45$
$$
\Rightarrow 29a \leqslant 45 \Rightarrow a = 1.
$$

Thus, $b + 4c = 29$
$$
\begin{array}{l}
\Rightarrow 0 \leqslant b = 29 - 4c \leqslant 9 \\
\Rightarrow -29 \leqslant -4c \leqslant -20 \\
\Rightarrow 5 \leqslant c \leqslant \frac{29}{4} \Rightarrow c \in \{5, 6, 7\}.
\end{array}
$$

When $c = 5$, $b = 9$, $A = 195$;
When $c = 6$, $b = 5$, $A = 156$;
When $c = 7$, $b = 1$, $A = 117$.