Three. (50 points) Let $n$ be a positive even number. Divide $\{1,2, \cdots, n^{2}\}$ into sets $A$ and $B$, satisfying $|A|=|B|$, and $S_{A}$, $S_{B}$ are the sums of the elements in sets $A$ and $B$, respectively. Find all $n$ such that there exists a partition satisfying $\frac{S_{A}}{S_{B}}=\frac{20}{19}$.
Three, $n=78 k\left(k \in \mathbf{N}_{+}\right)$ or $n=78 l+18$ $(l \in \mathbf{N})$ or $n=78 l+60(l \in \mathbf{N})$.
On one hand, it is obvious that $S_{A}$ and $S_{B}$ are both positive integers.
Thus, $20\left|S_{A}, 19\right| S_{B}$.
Let $S_{A}=20 T$. Then $S_{B}=19 T$.
By $S_{A}+S_{B}=1+2+\cdots+n^{2}=\frac{n^{2}\left(n^{2}+1\right)}{2}$
$\left.\Rightarrow 39 T=\frac{n^{2}\left(n^{2}+1\right)}{2} \Rightarrow 39 \right\rvert\, n^{2}\left(n^{2}+1\right)$.
Since $n^{2} \equiv 0$ or $1(\bmod 3)$, thus, $3 \nmid\left(n^{2}+1\right)$.
Noting that, $3\left|n^{2}\left(n^{2}+1\right) \Rightarrow 3\right| n$.
Also, $n$ is even, then $6 \mid n$, let $n=6 t\left(t \in \mathbf{Z}_{+}\right)$.
Hence $39\left|36 t^{2}\left(36 t^{2}+1\right) \Rightarrow 13\right| 12 t^{2}\left(36 t^{2}+1\right)$
$\Rightarrow 13 \mid t^{2}$ or $13 \mid\left(36 t^{2}+1\right)$.
If $13 \mid t^{2}$, then $13|t \Rightarrow 78| n$
$\Rightarrow n=78 k\left(k \in \mathbf{Z}_{+}\right)$.
If $13 \mid\left(36 t^{2}+1\right)$, then
$3 t^{2} \equiv 1 \equiv 27(\bmod 13) \Rightarrow t^{2} \equiv 9(\bmod 13)$
$\Rightarrow 13 \mid\left(t^{2}-9\right)$
$\Rightarrow 13 \mid(t-3)$ or $13 \mid(t+3)$.
Thus, $n=78 l+18(l \in \mathbf{N})$ or $n=78 l+60$ $(l \in \mathbf{N})$.

On the other hand, when $n=78 k\left(k \in \mathbf{N}_{+}\right)$ or $n=78 l+$ $18(l \in \mathbf{N})$ or $n=78 l+60(l \in \mathbf{N})$, let
$$
\begin{array}{l}
S=1+2+\cdots+n^{2}=\frac{n^{2}\left(n^{2}+1\right)}{2} . \\
\text { Let } A_{0}=\left\{1,2, \cdots, \frac{n^{2}}{2}\right\}, \\
A_{1}=\left\{\frac{n^{2}}{2}+1, \frac{n^{2}}{2}+2, \cdots, n^{2}\right\} .
\end{array}
$$

Then $S_{A_{0}}=1+2+\cdots+\frac{n^{2}}{2}$
$$
\begin{aligned}
= & \frac{1}{2} \times \frac{n^{2}}{2}\left(\frac{n^{2}}{2}+1\right)=\frac{n^{2}\left(n^{2}+2\right)}{8} \\
\Rightarrow & S_{A_{1}}=S-S_{A_{0}} \\
& =\frac{n^{2}\left(n^{2}+1\right)}{2}-\frac{n^{2}\left(n^{2}+2\right)}{8} \\
& =\frac{n^{2}\left(3 n^{2}+2\right)}{8} .
\end{aligned}
$$

To make $\frac{S_{A}}{S_{B}}=\frac{20}{19}$, it is only necessary to construct a subset $A$ of $\left\{1,2, \cdots, n^{2}\right\}$ such that $\frac{S_{A}}{S}=\frac{20}{39}$.
Noting that,
$$
\frac{S_{A_{0}}}{S}=\frac{n^{2}+2}{4\left(n^{2}+1\right)}\frac{20}{39},
$$

it can be achieved through a finite number of operations, each operation changing only one element and increasing the sum of elements by 1.

Thus, by the discrete intermediate value theorem, it is known that during this finite number of operations, there must be one operation that results in a set (denoted as $A$) that exactly satisfies $\frac{S_{A}}{S}=\frac{20}{39}$.