Four, (50 points) Find all positive odd integers $n$, such that there are an odd number of integers in $1,2, \cdots$, $\left[\frac{n}{4}\right]$ that are coprime with $n$.

保留源文本的换行和格式，直接输出翻译结果。
It is known that $n=1$ does not hold.
Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$.
Then, by the principle of inclusion-exclusion, the number of integers from 1 to $\left[\frac{n}{4}\right]$ that are coprime to $n$ is
$$
\begin{aligned}
S= & {\left[\frac{n}{4}\right]-\sum_{i=1}^{k}\left[\frac{\left[\frac{n}{4}\right]}{p_{i}}\right]+\sum_{1 \leqslant i<j \leqslant k}\left[\frac{\left[\frac{n}{4}\right]}{p_{i} p_{j}}\right]+\cdots+} \\
& (-1)^{r} \sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{i} \leqslant k}\left[\frac{\left[\frac{n}{4}\right]}{p_{i_{1}} p_{i_{2}} \cdots p_{i_{r}}}\right]+\cdots+ \\
& (-1)^{k}\left[\frac{\left[\frac{n}{4}\right]}{p_{1} p_{2} \cdots p_{k}}\right]
\end{aligned}
$$

If $2 \nmid n$, let
$\left[\frac{n}{4}\right]=\frac{n-\varepsilon}{4}(\varepsilon \in\{1,2,3\})$.
If $t \mid n$, and $t \geqslant 3$, then
$\left[\frac{\left[\frac{n}{4}\right]}{t}\right]=\left[\frac{\frac{n}{t}-\frac{\varepsilon}{t}}{4}\right]$,
$\frac{n}{\left[\frac{t}{4}\right]} \leqslant \frac{\frac{n}{t}-1}{4} \leqslant \frac{\frac{n}{t}-\frac{\varepsilon}{t}}{4}$.
Thus, $\left[\frac{\left[\frac{n}{4}\right]}{t}\right]=\frac{n}{\left[\frac{t}{4}\right]}=\left[\frac{n}{4 t}\right]$.
Therefore, $S=\sum_{r=0}^{k}(-1)^{r} \sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{r} \leqslant k}\left[\frac{n}{4 p_{i_{1}} p_{i_{2}} \cdots p_{i_{r}}}\right]$
$\equiv \sum_{r=0}^{k} \sum_{1 \leqslant i_{1}<i_{2}<\cdots<i, \leqslant k}\left[\frac{n}{4 p_{i_{1}} p_{i_{2}} \cdots p_{i_{r}}}\right](\bmod 2)$.
Moreover, when $M$ is odd,
$\left[\frac{M}{4}\right] \equiv 1(\bmod 2)$
$\Leftrightarrow M \equiv 5(\bmod 8)$ or $M \equiv 7(\bmod 8)$.
If there exists $p_{i_{4}} \equiv 1(\bmod 8)$, then by equation (1), for any $a \in \mathbf{Z}_{+}$, we have
$$
\left[\frac{a}{4}\right]+\left[\frac{p_{i_{t}} a}{4}\right] \equiv 0(\bmod 2) .
$$

Thus, $S$
$$
\begin{array}{l}
\equiv \sum_{r=0}^{k} \sum_{\substack{1 \leqslant i_{1}<i_{2}<\cdots<i_{i} \leq k \\
i_{1} \neq i_{i}}}\left(\left[\frac{n}{4 p_{i_{1}} p_{i_{2}} \cdots p_{i_{r}}}\right]+\left[\frac{n}{4 p_{i_{1}} p_{i_{2} \cdots p_{i}, p_{i}}}\right]\right) \\
\equiv 0(\bmod 2) , \\
\end{array}
$$

which is a contradiction.
Similarly, if $p_{i_{i}} \equiv 3(\bmod 8)$, we also have
$$
S \equiv 0(\bmod 2),
$$

which is a contradiction.
Therefore, the prime factors of $n$ can only be of the form $8 k+5$ or $8 k+7$.

Assume that $n$ has $s$ prime factors of the form $8 k+5$ and $t$ prime factors of the form $8 k+7$.

If $p \equiv 5$ or $7(\bmod 8)$, then by equation (1), for any $a \in \mathbf{Z}_{+}$, we have
$$
\left[\frac{a}{4}\right]+\left[\frac{p a}{4}\right] \equiv 0(\bmod 2) .
$$

Thus, $S$
$$
\begin{array}{l}
\equiv \sum_{r=0}^{k-1} \sum_{2 \leqslant i_{1}<i_{2}<\cdots<i_{1} \leqslant k}\left(\left[\frac{n}{4 p_{i_{1}} p_{i_{2}} \cdots p_{i_{i}}}\right]+\left[\frac{n}{4 p_{1} p_{i_{1}} \cdots p_{i_{i}}}\right]\right) \\
\equiv \sum_{r=0}^{k-1} \sum_{2 \leqslant i_{1}<i_{2}<\cdots<i_{i} \leqslant k} 1 \\
=\mathrm{C}_{k-1}^{0}+\mathrm{C}_{k-1}^{1}+\cdots+\mathrm{C}_{k-1}^{k-1} \\
=2^{k-1}(\bmod 2) . \\
\end{array}
$$

From $2+S$, we get $k=1$.
Therefore, $S=\left[\frac{n}{4}\right]-\left[\frac{n}{4 p_{1}}\right] \equiv 1(\bmod 2)$, which satisfies the problem's condition.
In conclusion, $n=p^{\alpha}$, where
$\alpha \in \mathbf{Z}_{+}, p \equiv 5$ or $7(\bmod 8)$.