Example 2 (2007 Turkey National Team Selection Exam) Find all positive odd integers $n$, such that there exists a positive odd integer

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Example 2 (2007 Turkey National Team Selection Exam) Find all positive odd integers $n$, such that there exists a positive odd integer
Since $n$ is a positive odd number, we have
$n^{4} \equiv 1(\bmod 8)$.
Also, since $x_{i}(1 \leqslant i \leqslant n)$ are positive odd numbers, it follows that $x_{i}^{2} \equiv 1(\bmod 8)$.
Therefore, $n \equiv x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \equiv n^{4} \equiv 1(\bmod 8)$.
On the other hand, if $n \equiv 1(\bmod 8)$, then there exist $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy the conditions. If $n=1$, let $x_{1}=1$, then $n^{4}=1=x_{1}^{2}$.
If $n=8 k+1\left(k \in \mathbf{N}_{+}\right)$, then
$$
\begin{aligned}
n^{4}= & (8 k+1)^{4} \\
= & (8 k-1)^{4}+(8 k+1)^{4}-(8 k-1)^{4} \\
= & (8 k-1)^{4}+\left[(8 k+1)^{2}-(8 k-1)^{2}\right] \cdot\left[(8 k+1)^{2}+(8 k-1)^{2}\right] \\
= & (8 k-1)^{4}+32 k\left(128 k^{2}+2\right) \\
= & (8 k-1)^{4}+4 k(32 k-1)^{2}+(16 k-1)^{2}+(92 k-1) \\
= & (8 k-1)^{4}+4 k(32 k-1)^{2}+(16 k-1)^{2}+92(k-1)+91 \\
= & (8 k-1)^{4}+4 k(32 k-1)^{2}+(16 k-1)^{2}+(k-1)\left(9^{2}+3^{2}+1^{2}+1^{2}\right)+ \\
& \left(9^{2}+3^{2}+1^{2}\right) .
\end{aligned}
$$

Therefore, $n^{4}$ can be expressed as the sum of the squares of $1+4 k+1+4(k-1)+3=8 k+1=n$ odd numbers.
In conclusion, the result is $n=8 k+1(k \in \mathbf{N})$.