Three. (25 points) A positive integer $M$ has 8 positive divisors. Xiao Feng, while calculating the sum of the 8 positive divisors of $M$, forgot to add one of the divisors, resulting in a sum of 2776. Find all possible values of the positive integer $M$.


Three, if $M$ has at least four different prime factors, then the number of positive divisors of $M$ is at least $(1+1)^{4}=16$, which contradicts the problem statement.
Therefore, $M$ has at most three different prime factors.
If all prime factors of $M$ are odd, then all positive divisors of $M$ are odd. In this case, the sum of the seven positive divisors of $M$ is odd, which contradicts the fact that the sum is 2776, an even number.
Therefore, one of the prime factors of $M$ is 2.
(1) When $M$ has only one prime factor, this prime factor is $2$, and $M$ is $2^{7}$. Let the missing positive divisor be $2^{n}(n \in \mathbf{N}, n \leqslant 7)$.
By the problem statement,
$1+2+2^{2}+\cdots+2^{7}-2^{n}=2776$.
Simplifying, we get $2^{n}=-2521$ (discard).
Therefore, $M \neq 2^{7}$.
Thus, $M$ cannot have only one prime factor.
(2) When $M$ has only two different prime factors, let these two different prime factors be $p$ and $q$. Then $M$ can be set as $p^{3} q$. The missing positive divisor is $p^{m} q^{n}(m, n \in \mathbf{N}, 0 \leqslant m \leqslant 3, 0 \leqslant n \leqslant 1$.
By the problem statement,
$$
\left(1+p+p^{2}+p^{3}\right)(1+q)-p^{m} q^{n}=2776 \text {. }
$$
(i) If $p=2$, then equation (1) becomes
$$
15 q-2^{m} \times q^{n}=2761 \text {. }
$$

After calculation, only when $(m, n)=(2,1)$, $q=251$ (a prime number) meets the problem's requirements.
At this point, $M=2^{3} \times 251=2008$.
(ii) If $q=2$, then equation (1) becomes
$3\left(p+p^{2}+p^{3}\right)-2^{n} \times p^{m}$
$$
=2773=47 \times 59 \text {. }
$$

When $m \geqslant 1$, the left side of equation (2) can be divided by $p$, then $p \mid 47 \times 59$.
Since $p$ is a prime factor of $M$, hence $p=47$ or 59.
Therefore, the left side of equation (1) $\geqslant p^{3} \geqslant 47^{3}>2776$.
So, $m \geqslant 1$ does not hold.
When $m=0$, equation (2) becomes
$3 p\left(1+p+p^{2}\right)=2773+2^{n}=2775$ or 2774.
Since $3 \nmid 2775, 3 \nmid 2774$, so,
$3 p\left(1+p+p^{2}\right)=2775$,
$p\left(1+p+p^{2}\right)=925=5^{2} \times 37$.
Therefore, $p \mid 5^{2} \times 37$, hence $p=5$ or 37.
When $p=5$ or 37, equation (3) does not hold.
So, $q \neq 2$.
(3) When $M$ has only three different prime factors, one of them is 2, and the other two are denoted as $p$ and $q(p \neq q)$.
By the problem statement, $M=2 p q$.
Let the missing positive divisor be $2^{m} p^{n} q^{t}(m, n, t$ are natural numbers not exceeding 1).
By the problem statement,
$$
(1+2)(1+p)(1+q)-2^{m} p^{n} q^{t}=2776,
$$

i.e., $3(1+p)(1+q)-2^{m} p^{n} q^{t}=2776$.
Since $p, q$ are odd prime numbers, therefore,
$4 \mid 3(1+p)(1+q)$.
Also, $4 \mid 2776$, by equation (4), $4 \mid 2^{m} p^{n} q^{t}$.
Hence, $4 \mid 2^{m}$.
Since $m \leqslant 1$, then $4 \nmid 2^{m}$. Therefore, equation (5) does not hold.
Thus, $M$ cannot have three different prime factors.
In summary, the value of $M$ is 2008.
(Xie Wenxiao, Huanggang High School, Hubei Province, 438000)