1. (5 points) Arrange 6 consecutive natural numbers in ascending order. If the sum of the last 3 numbers is twice the sum of the first 3 numbers, then the largest number among these 6 numbers is $\qquad$, and the sum of these 6 numbers is $\qquad$.
【Analysis】We can first set any one of the natural numbers as $n$, express the other natural numbers, and solve the problem by setting up an equation according to the problem statement.

【Solution】Solution: Let the 6 consecutive natural numbers be $n-2, n-1, n, n+1, n+2, n+3$
$$
\begin{array}{l}
{[(n-2)+(n-1)+n] \times 2=(n+1)+(n+2)+(n+3) \text {, }} \\
(3 n-3) \times 2=3 n+6 \text {, } \\
n=4 \text {; } \\
\end{array}
$$

So the 6 consecutive natural numbers are $2,3,4,5,6,7$,
The largest of the 6 numbers is 7,
The sum of these 6 numbers is 27,
Therefore, the answers are: 7,27.