1013. Find the two smallest consecutive natural numbers, each of which has a sum of digits divisible by 7.
$\triangle$ Obviously, this is possible only with "skipping" over the nine.

Let the smaller of the sought numbers be denoted by

$$
x=\overline{a b c \ldots k l 99 \ldots 9}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_475fdfbb473422773b9eg-223.jpg?height=48&width=666&top_left_y=1004&top_left_x=341)

$$
x+1=\overline{a b c \ldots k(l+1) 00 \ldots 0}
$$

( $i$ zeros)

By the condition, each of the sums of the digits

$$
a+b+c+\ldots+k+l+9 n, \quad a+b+c+\ldots+k+(l+1)
$$

is divisible by 7. Therefore, the difference between these sums is also divisible by 7:

$$
(9 n-1): 7
$$

The smallest value of $n$ that satisfies the last relation is $n=4$. Then $9 n=36$.

For the number $\overline{a b c . . . k l}$, we need to take a single-digit number, the sum of which with 36 is divisible by 7. Clearly, this single-digit number is 6. Therefore,

$$
x=69999, \quad x+1=70000
$$

Answer: $69999,70000$.