9.042. $0.3^{2+4+6+\ldots+2x}>0.3^{72}(x \in N)$.
## Solution.

The given inequality is equivalent to the inequality $2+4+6+\ldots+2x<72$, where the left side of the inequality is the sum of the terms of an arithmetic progression with $a_{1}=2, d=2, a_{n}=2x$. Then we get

$$
\left\{\begin{array}{l}
\frac{2+2x}{2} \cdot n < 72 \\
n = \frac{2x-2}{2} + 1
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(1+x) x < 72 \\
n = x
\end{array}\right.\right.
$$

From this, we have $x^2 + x - 72 < 0, -9 < x < 8$. Since $x = n \in \mathbb{N}$, then $n=1,2,3,4,5,6,7$.

Answer: $\quad 1,2,3,4,5,6,7$.