Four. (50 points) Let $M$ be a set composed of $n\left(n \in \mathbf{N}_{+}\right)$ distinct positive integers, where each element has prime factors no greater than 100, and there do not exist four distinct elements in $M$ such that the product of these four numbers is a fourth power. Find the maximum value of $n$.

 untranslated part:
(50 points)设 $M$ 是由 $n\left(n \in \mathbf{N}_{+}\right)$个不同的正整数组成的集合,其中每个元素的质因子不大于 100 , 且 $M$ 中不存在四个不同的元素,使得这四个数之积是一个 4 次方数.求 $n$的最大值.

The translated part:
(50 points) Let $M$ be a set composed of $n\left(n \in \mathbf{N}_{+}\right)$ distinct positive integers, where each element has prime factors no greater than 100, and there do not exist four distinct elements in $M$ such that the product of these four numbers is a fourth power. Find the maximum value of $n$.
Prime numbers not greater than 100 are
$2,3,5,7,11,13,17,19,23,29,31,37,41$, $43,47,53,59,61,67,71,73,79,83,89,97$.

Denoted as $p_{i}(i=1,2, \cdots, 25)$. $\alpha_{i} \in \mathbf{N}(i=1,2, \cdots, 25)$.

The tuple $\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{25}\right)$, classified by parity, has $2^{25}$ types. Thus, from $M$, we can select
$$
\left[\frac{n-2^{25}-1}{2}\right]+1=\left[\frac{n-1}{2}\right]-2^{24}+1
$$

For the tuples $\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{25}\right)$ and $\left(\alpha_{1}^{\prime}, \alpha_{2}^{\prime}, \cdots, \alpha_{25}^{\prime}\right)$,
if they are of the same type, then
$$
\alpha_{i}+\alpha_{i}^{\prime}=2 \beta_{i}\left(\beta_{i} \in \mathbf{N}, i=1,2, \cdots, 25\right) .
$$

Thus, we obtain $\left[\frac{n-1}{2}\right]-2^{24}+1$ tuples $\left(\beta_{1}, \beta_{2}, \cdots, \beta_{25}\right)$.

If $\left[\frac{n-1}{2}\right]-2^{24}+1 \geqslant 2^{25}+1$, then we can select a pair of tuples $\left(\beta_{1}, \beta_{2}, \cdots, \beta_{25}\right)$ and $\left(\beta_{1}^{\prime}, \beta_{2}^{\prime}, \cdots, \beta_{25}^{\prime}\right)$ from the above tuples, which are of the same type, i.e.,
$$
\beta_{i}+\beta_{i}^{\prime}=2 \gamma_{i}\left(\gamma_{i} \in \mathbf{N}, i=1,2, \cdots, 25\right) .
$$

The product of the 4 elements in $M$ corresponding to these tuples is $\prod_{i=1}^{25} p_{i}^{4 \gamma_{i}}$, which is a fourth power number. Therefore,
$$
\left[\frac{n-1}{2}\right]-2^{24}+1 \leqslant 2^{25} \Rightarrow\left[\frac{n+1}{2}\right] \leqslant 3 \times 2^{24} .
$$

Thus, $n_{\max } \leqslant 3 \times 2^{25}$.
Next, we prove: $n_{\text {max }}=3 \times 2^{25}$.
For each component of each type, if it is odd, take it as $4 k+1$; if it is even, take it as $4 k(k=0,1,2)$.

Thus, each type of tuple has 3, totaling $3 \times 2^{25}$ tuples.

We will prove by contradiction that: among the above tuples, there do not exist 4 tuples such that the sum of all components is a multiple of 4.
Assume there exist 4 tuples
$$
\begin{array}{l}
\left(a_{1}, a_{2}, \cdots, a_{25}\right),\left(b_{1}, b_{2}, \cdots, b_{25}\right), \\
\left(c_{1}, c_{2}, \cdots, c_{25}\right),\left(d_{1}, d_{2}, \cdots, d_{25}\right),
\end{array}
$$

such that each component $a_{i} 、 b_{i} 、 c_{i} 、 d_{i} \equiv 0$ or $1(\bmod 4)$ and $a_{i}+b_{i}+c_{i}+d_{i} \equiv 0(\bmod 4)(i=1,2, \cdots, 25)$.
It is easy to see that $a_{i} \equiv b_{i} \equiv c_{i} \equiv d_{i} \equiv 0$ or $1(\bmod 4)$.
Thus, $\left(a_{1}, a_{2}, \cdots, a_{25}\right) \equiv\left(b_{1}, b_{2}, \cdots, b_{25}\right)$
$\equiv\left(c_{1}, c_{2}, \cdots, c_{25}\right)$
$$
\equiv\left(d_{1}, d_{2}, \cdots, d_{25}\right)(\bmod 4) .
$$

Therefore, the above 4 tuples are of the same type. But each type has only 3 tuples, which is a contradiction.
Hence, $n_{\text {max }}=3 \times 2^{25}$.