3. Among all possible four-digit numbers formed using the digits $1,9,9,0$, for each such four-digit number and a natural number $n$, their sum when divided by 7 does not leave a remainder of 1. List all such natural numbers $n$ in descending order.
$$
n_{1}<n_{2}<n_{3}<n_{4}<\cdots \cdots,
$$

Find: the value of $n_{1}$.
3. Solution
$1,0,9,0$ digits can form the following four-digit numbers: 1099, 1909, 1990, 9019, 9091, 9109, $9190,9901,9910$ for a total of nine. The remainders when these nine numbers are divided by 7 are $0,5,2,3,5,2,6$, 3 , 5 . Since $n$ and their sum cannot be divisible by 7 with a remainder of 1, $n$ cannot have a remainder of $1,3,6,5$ : $3,6,2,5,3$ when divided by 7. That is, $n$ cannot have a remainder of $1,2,3,5,6$ when divided by 7. Therefore, the remainder when $n$ is divided by 7 can only be 0 or 4. Since $n$ is a natural number, $n_{1}=4, n_{2}=7$, $n_{1} n_{2}=28$.