Let $x=\overline{a_{n} a_{n-1}...a_{0}}=10^{n} a_{n}+10^{n-1}+...+a_{0}$ be all positive integers which are equal to $13$ times the sum of their digits. Then $10^{n} a_{n}+10^{n-1}+...+100a_{2}+10a_{1}+a_{0}=13a_{n}+13a_{n-1}+...+13a_{2}+13a_{1}+13a_{0} \Rightarrow$ $\Rightarrow (10^{n}-13)a_{n}+(10^{n-1}-13)a_{n-1}+...+87a_{2}=3a_{1}+12a_{0}$. - If $n \ge 4$, then $LHS \ge 987>135=15 \cdot 9=3\cdot 9+12 \cdot 9 \ge 3a_{1}+12a_{0}=RHS$, contradiction. - If $n=3$, then let $\overline{abc}=100a+10b+c$ be a three-digit number which is equal to $13$ time the sum of its digits. Then: $87a=3b+12c \Rightarrow 29a=b+4c \le 5 \cdot 9=45 \Rightarrow a\le 2 \Rightarrow a=1 \Rightarrow b+4c=29$, which gives $(b,c)=(1,7),(5,6),(9,5)$. - If $n=2$, then let $\overline{ab}=10a+b$ be a two-digit number which is equal to $13$ time the sum of its digits. Then $3a+12b=0 \Rightarrow a+4b=0$, contradiction. Conclusively the asked numbers are $117,156,195$.

Mod 9 can not be used here?