16.1.7 $7 \star$ Find all positive integers $N$, such that it equals the sum of the digits of $N^{3}$.
Let $N$ be a $k$-digit number. $N^{3}$ has at most $3k$ digits, and its digit sum $\leqslant 9 \times 3k = 27k$. When $k \geqslant 3$,
$$
N \geqslant 10^{k-1} = 10 \times (1+9)^{k-2} > 10 \times 9(k-2) > 27k,
$$

Therefore, $N$ is a one-digit or two-digit number. $N^{3}$ has at most 6 digits, and the digit sum $\leqslant 6 \times 9 = 54$. Thus, $N \leqslant 54$. The digit sum $S \equiv N \equiv N^{3} \equiv 0, \pm 1 (\bmod 9)$. Therefore, $S = 1, 8, 9, 10$. $N = 1, 8$, $9, 10, 17, 18, 19, 26, 27, 28, 35, 36, 37, 44, 45, 46, 53, 54$.
Upon verification, the required $N$ is $1, 8, 17, 18, 26, 27$.