Initial 26. Find all natural numbers of the form $\overline{30 x 070 y 03}$ that are divisible by 37.
Solve $\overline{30 x 070 y 03}=300070003+10^{6} x+10^{2} y$
$$
\begin{aligned}
= & 37(8110000+27027 x+2 y) \\
& +(3+x-11 y) .
\end{aligned}
$$

Therefore, from the given conditions, $3+x-11 y$ is a multiple of 37.
And $0 \leqslant x, y \leqslant 9, -96 \leqslant 3+x-11 y \leqslant 12$. Thus, the value of $3+x-11 y$ can only be $0, -37$ or -74.
(1) If $3+x-11 y=0$, then by $0 \leqslant x, y \leqslant 9$ we get $x=8, y=1$.
(2) If $3+x-11 y=-37$, the integer solution for this equation within 0 to 9 is $x=4, y=4$.
(3) If $3+x-11 y=-74$, then the corresponding values of $x, y$ are $x=0, y=7$.
In summary, the required natural numbers are 3080701 C, 304070403 or 300070703.
(Liu Jinyi, Hebei Yongqing Normal School, 3026:0)