We distinguish different cases for $n$ based on the remainder of $n$ when dividing by $3$.

If $n = 3k$, then the grasshopper can jump as follows:
$1; 3, 6, \ldots, 3(k-1); 3(k-1)+2, 3(k-2)+2, \ldots, 2; 4, 7, \ldots, 3(k-1)+1; 3k$.
The grasshopper jumps over triples on the way out, over triples plus $2$ on the way back, and then over triples plus $1$ on the second way out.

If $n = 3k + 1$, then the grasshopper can jump as follows:
$1; 3, 6, \ldots, 3k; 3k-2, 3(k-1)-2, \ldots, 7, 4; 2, 5, 8, \ldots, 3(k-1)+2; 3k+1$.

Finally, for $n = 3k + 2$, we find that the grasshopper can jump as follows:
$1, 3, 6, 4; 2, 5, 8, \ldots, 3(k-1)+2; 3k+1, 3(k-1)+1, \ldots, 7; 9, \ldots, 3k; 3k+2$.

Because $n \ge 9$ we have that $k \ge 3$. In the solutions above for $k \ge 3$ we see that every leg of the 'zigzag' is nonempty, so a solution exists.