For $n>2$, we have
$$
\begin{aligned}
55 \underbrace{99 \cdots 989}_{n-2} \underbrace{33 \cdots 39}_{*-1} & =55 \cdot 10^{2 n}+\left(10^{2 n}-10^{n+2}\right)+89 \cdot 10^{n}+\frac{1}{3}\left(10^{n}-10\right)+9 \\
& =\frac{1}{9}\left(504 \cdot 10^{2 n}-96 \cdot 10^{n}+51\right) .
\end{aligned}
$$

Let positive integers $u, v, w$ and integers $x, y, z\left(|x| 、|y| 、|z|<10^{n}\right)$, such that
$$
\left(10^{n} u+x\right)^{2}+\left(10^{n} v+y\right)^{2}+\left(10^{n} w+z\right)^{2}=504 \cdot 10^{2 n}-96 \cdot 10^{n}+51 \text {. }
$$

By comparing the coefficients of $10^{2 n}, 2 \cdot 10^{n}, 1$, we get
$$
\begin{array}{c}
u^{2}+v^{2}+w^{2}=504=22^{2}+4^{2}+2^{2}=20^{2}+10^{2}+2^{2}=18^{2}+12^{2}+6^{2}, \\
u x+v y+w z=-48
\end{array}
$$

and $x^{2}+y^{2}+z^{2}=51=7^{2}+1^{1}+1^{2}=5^{2}+5^{2}+1^{2}$.
$(u, v, w)$ can only be $(22,4,2),(20,10,2)$ or $(18,12,6)$, combined with the possible values of $(x, y, z)$ $( \pm 7, \pm 1, \pm 1),( \pm 5, \pm 5, \pm 1)$ and equation (3), we get the following table:
\begin{tabular}{rrrrrr}
$u$ & $v$ & $w$ & $x$ & \multicolumn{1}{c}{$\boldsymbol{y}$} & \multicolumn{1}{c}{$\boldsymbol{z}$} \\
\hline 22 & 4 & 2 & -1 & -7 & 1 \\
20 & 10 & 2 & 1 & -7 & 1 \\
20 & 10 & 2 & -5 & 5 & 1 \\
18 & 12 & 6 & -1 & 1 & -7 \\
18 & 12 & 6 & -5 & 1 & 5 \\
18 & 12 & 6 & -1 & -5 & 5
\end{tabular}

From (1) and (2), we need $\frac{10^{n} u+x}{3}, \frac{10^{n} v+y}{3}, \frac{10^{n} w+z}{3}$ to be integers, so $u+x, v+y$, and $w+z$ must all be divisible by 3. Thus, only the first three rows in the table satisfy the condition. The given number can be expressed in the following three ways:
$$
\begin{array}{c}
7 \underbrace{33 \cdots 3^{2}}_{n}+1 \underbrace{33 \cdots 3}_{n-1} 1^{2}+\underbrace{66 \cdots 6}_{n-1} 7^{2}, \\
\underbrace{66 \cdots 6}_{n} 7^{2}+\underbrace{33 \cdots 3}_{n} 1^{2}+\underbrace{66 \cdots 6}_{n} 7^{2}, \\
\underbrace{66 \cdots 6}_{n} 5^{2}+\underbrace{33 \cdots 3}_{n} 5^{2}+\underbrace{66 \cdots 6}_{n} 7^{2} .
\end{array}
$$
19.7.11 ** For each positive integer $n>2$, the number $55 \underbrace{99 \cdots 9}_{n-2 \uparrow 9} 89 \underbrace{33 \cdots 39}_{n-1 \uparrow 3}$ can be expressed as the sum of three positive integers squared.