11 is not a good number. Because 11 can only be added to 5 to get a square number, and 4 must be added to 5 to get a square number. Therefore, at least one of $11+a_{11}$ and $4+a_{4}$ is not a square number.
13 is a good number, because
$$8,2,13,12,11,10,9,1,7,6,5,4,3$$

is a valid arrangement.
15 is a good number, because
$$15,14,13,12,11,10,9,8,7,6,5,4,3,2,1$$

is a valid arrangement.
17 is a good number, because
$$3,7,6,5,4,10,2,17,16,15,14,13,12,11,1,9,8$$

is a valid arrangement.
19 is a good number, because
$$8,7,6,5,4,3,2,1,16,15,14,13,12,11,10,9,19,18,17$$

is a valid arrangement.
Note: It can be proven that $1,2,4,6,7,11$ are not good numbers, and the rest of the natural numbers are good numbers. The proof uses induction. It is not difficult to verify that the conclusion holds for $n \leqslant 25$.
Assume that the conclusion holds for $n \leqslant(m-1)^{2}(m \geqslant 6)$. Consider $(m-1)^{2}+j$, where the natural number $j \leqslant 2 m-1$.
Since $m \geqslant 6$,
$$\begin{array}{c} (m+1)^{2}-\left((m-1)^{2}+j\right)=4 m-j<4 m<(m-1)^{2}, \\ 4 m-j-1 \geqslant 2 m \geqslant 12, \end{array}$$

so $1,2, \cdots, 4 m-j-1=h$ can be arranged into $a_{1}, a_{2}, \cdots, a_{h}$, such that $k+a_{k}(k=1,2, \cdots, h)$ are all square numbers. And $4 m-j, 4 m-j+1, \cdots,(m-1)^{2} 1-j$ can be arranged in reverse order,
$$(4 m-j+i)+\left((m-1)^{2}+j-i\right)=(m+1)^{2},$$

so $1,2, \cdots, 4 m-j-1,4 m-j, \cdots,(m-1)^{2}+j$ can be arranged as
$$a_{1}, a_{2}, \cdots, a_{h},(m-1)^{2}+j, \cdots, 4 m-j,$$

satisfying $k+a_{k}\left(k=1,2, \cdots,(m-1)^{2}+j\right)$ are all square numbers.
Therefore, all numbers greater than 11 are good numbers.