Solution:

Suppose $N$ has base-$7$ digits $d_k \ldots d_1 d_0.$ The condition is $\sum d_i \, 10^i = 2 \sum d_i \, 7^i,$ that is $\sum d_i \,(10^i - 2 \cdot 7^i) = 0.$ The coefficients $10^i - 2 \cdot 7^i$ for $i = 0, 1, 2, 3$ are $-1,$ $-4,$ $2,$ $314.$ If there were a digit $d_3 \ge 1,$ the positive contribution would be at least $314,$ but the negative terms total at most $4 \cdot 6 + 6 = 30.$ So $N$ has at most three base-$7$ digits.

For three digits the condition reads $2 d_2 = 4 d_1 + d_0.$ To maximize $N = 49 d_2 + 7 d_1 + d_0,$ take $d_2 = 6,$ so $4 d_1 + d_0 = 12;$ the largest value of $7 d_1 + d_0$ comes from $d_1 = 3,$ $d_0 = 0.$

Thus $N = 49 \cdot 6 + 7 \cdot 3 = 315,$ whose base-$7$ representation is $630 = 2 \cdot 315.$