Three, (50 points) Try to find the smallest positive integer $M$, such that the sum of all positive divisors of $M$ is 4896.
Let $M=\prod_{i=1}^{n} p_{i}^{\alpha_{i}}\left(p_{i}\right.$ be a prime, $\alpha_{i} \in \mathbf{N}_{+}, i=1,2, \cdots, n)$. Denote
$$
f(p, \alpha)=\sum_{k=0}^{\alpha} p^{k}\left(p\right.$ be a prime, $\left.\alpha \in \mathbf{N}_{+}\right)$.
From the problem, we know
$$
\prod_{i=1}^{n} f\left(p_{i}, \alpha_{i}\right)=4896=2^{5} \times 3^{2} \times 17.
$$

If $n=1$, then $f\left(p_{1}, \alpha_{1}\right)=4896$, which implies
$$
p_{1} \mid 4895.
$$

Since $4895=5 \times 11 \times 89 \Rightarrow p_{1} \in\{5,11, 89\}$.
$$
\begin{array}{l}
\text { But } f(5,5)=1.
\end{array}
$$
Next, we divide the positive divisors of 4896 into the following two sets:
$$
\begin{aligned}
A= & \{1,2,9,16,17,34,36,51,96,136, \\
& 144,153,204,272,288,306,408, \\
& 544,612,816,1632,4896\}, \\
B= & \{3,4,6,8,12,18,24,32,48,68,72, \\
& 102,1224,2448\}.
\end{aligned}
$$

Similarly, we know: when $t \in A$, $f(p, \alpha)=t$ has no prime solution $p$; when $t \in B$, $f(p, \alpha)=t$ has only the solution
$$
(p, \alpha)=(t-1,1).
$$

From equation (1), assume $17 \mid f\left(p_{1}, \alpha_{1}\right)$. Then
$$
f\left(p_{1}, \alpha_{1}\right) \in\{68,102,1224,2448\}.
$$
(1) If $f\left(p_{1}, \alpha_{1}\right)=2448$, then
$$
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=2,
$$

which has no solution.
(2) If $f\left(p_{1}, \alpha_{1}\right)=1224$, then
$$
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=4.
$$

Thus, $M_{1}=1223 \times 3=3669$.
(3) If $f\left(p_{1}, \alpha_{1}\right)=102$, then
$$
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=48=4 \times 12=6 \times 8.
$$

Thus, $M_{2}=101 \times 48=4747$,
$$
\begin{array}{l}
M_{3}=101 \times 3 \times 11=3333, \\
M_{4}=101 \times 5 \times 7=3535 ;
\end{array}
$$
(4) If $f\left(p_{1}, \alpha_{1}\right)=68$, then
$$
\begin{array}{l}
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=72=3 \times 24 \\
=4 \times 18=6 \times 12=3 \times 4 \times 6.
\end{array}
$$

Thus, $M_{5}=67 \times 71=4757$,
$$
\begin{array}{l}
M_{6}=67 \times 2 \times 23=3082, \\
M_{7}=67 \times 3 \times 17=3417, \\
M_{8}=67 \times 5 \times 11=3685, \\
M_{9}=67 \times 2 \times 3 \times 5=2010.
\end{array}
$$

In summary, $M_{\min }=2010$.