Five. (15 points) If for any $n$ consecutive positive integers, there always exists a number whose sum of digits is a multiple of 8. Determine the minimum value of $n$. And explain the reason.

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Five, first prove that when $n \leqslant 14$, the property in the problem does not hold.
When $n=14$, for
$9999993, 9999994, \cdots, 10000006$
these 14 consecutive integers, the sum of the digits of any number cannot be divisible by 8.
Therefore, when $n \leqslant 14$, the property in the problem does not hold.
Thus, to make the property in the problem hold, we should have
$n \geqslant 15$.
Next, prove that when $n=15$, the property in the problem holds.
Let $a_{1}, a_{2}, \cdots, a_{15}$ be any 15 consecutive positive integers. Then, among these 15 positive integers, the number of integers with a units digit of 0 is at most two and at least one, which can be divided into:
(1) When $a_{1}, a_{2}, \cdots, a_{15}$ have two integers with a units digit of 0, let $a_{i}<a_{j}$, and $a_{i}, a_{j}$ have a units digit of 0, then $a_{i}, a_{i}+1, \cdots, a_{i}+9, a_{j}$ are 11 consecutive integers, among which $a_{i}, a_{i}+1, \cdots, a_{i}+9$ have no carry.

Let $n_{i}$ represent the sum of the digits of $a_{i}$, then the sum of the digits of the first 10 numbers are $n_{i}, n_{i}+1, \cdots, n_{i}+9$. Therefore, among these 10 consecutive numbers, at least one is divisible by 8.
(2) When $a_{1}, a_{2}, \cdots, a_{15}$ have only one integer with a units digit of 0 (denoted as $a_{i}$):
(i) If the integer $i$ satisfies $1 \leqslant i \leqslant 8$, then there are at least 7 consecutive integers after $a_{i}$. Thus, $a_{i}, a_{i}+1, \cdots, a_{i}+7$ are 8 consecutive integers, and the sum of the digits of these 8 consecutive integers are also 8 consecutive integers. Therefore, there must be one number that is divisible by 8.
(ii) If the integer $i$ satisfies $9 \leqslant i \leqslant 15$, then there are at least 8 consecutive integers before $a_{i}$, which can be denoted as $a_{i-8}$, $a_{i-7}, \cdots, a_{i-1}$. The sum of the digits of these 8 consecutive integers are also 8 consecutive integers. Therefore, there must be one number that is divisible by 8.

In summary, for any 15 consecutive integers, there must be one number whose sum of digits is a multiple of 8.

And for any fewer than 15 consecutive integers, this property does not hold.
Therefore, the minimum value of $n$ is 15.