Three, (50 points) Given $m \in \mathbf{N}_{+}$, let $f(a)$ denote the sum of the digits of the positive integer $a$. Try to find the minimum value of the positive integer $t$, such that among any $t$ consecutive positive integers, there is always a number $c$ satisfying $m \mid f(c)$.

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Let $k=\left[\frac{m-1}{9}\right], r=m-9 k$. Then,
when $10 \times(a+1)$, $f(a+1)=f(a)+1$;
when $10^{i} \|(a+1)\left(i \in \mathbf{N}_{+}\right)$,
$$
f(a+1)=f(a)+1-9 i=f\left(a+1-10^{i}\right)+1 \text {. }
$$

First, we prove: In any $t=2 r \cdot 10^{k}-1$ consecutive positive integers, there is always a number $c$ such that $m \mid f(c)$.

If these $t$ numbers include a multiple of $10^{k+1}$, let it be $a=10^{i} b(i \geqslant k+1,10 \nmid b)$.
Then $f(a)=f(b)=s$.
By $f\left(a+10^{\circ} u+10^{b}-1\right)=s+u+9 v$,

where $u=0,1, \cdots, 8 ; v=0,1, \cdots, k$, we know that the $f$ values of $\{a, a+$ $\left.1, \cdots, a+10^{k} r-1\right\}$ are
$$
F_{1}=\{s, s+1, \cdots, s+9 k+r-1\},
$$

Among these, there must be one that is a multiple of $9 k+r$.
By $f\left(a-10^{\circ} u\right)=s+9 i-(9 v+u)$, where $u=0,1, \cdots, 9 ; v=0,1, \cdots, k$, we know that the $f$ values of $\{a-1, a$ $\left.-2, \cdots, a-10^{k} r\right\}$ are
$$
F_{2}=\{s+9 i-1, s+9 i-2, \cdots, s+9 i-(9 k+r)\},
$$
Among these, there must be one that is a multiple of $9 k+r$.

If these $t$ numbers do not include a multiple of $10^{k+1}$, then they include a multiple of $10^{k}$. Let the smallest multiple be $a, f(a)=s$.

Similarly, we know that the $f$ values of $\left\{a, a+1, \cdots, a+10^{k} r-1\right\}$ are $F_{1}$, among which, there must be one that is a multiple of $9 k+r$.

In summary, in $2 r \cdot 10^{k}-1$ consecutive positive integers, there must be a number $c$ such that $m \mid f(c)$.
Thus, $t_{\min } \leqslant 2 r \cdot 10^{k}-1$.
Next, we consider three cases to find $t_{\text {min }}$.
(1) When $r=1,2,4,5,7,8$, $t_{\min }=2 r \cdot 10^{k}-1$.
Below, we construct $2 r \cdot 10^{k}-2$ consecutive positive integers such that none of them satisfy $m \mid f(c)$.
Take $a=10^{i}$, satisfying $(9 k+r) \mid(9 i+1)$. Then in
$$
A=\left\{a-10^{k} r, a-10^{k} r+1, \cdots, a+10^{k} r-1\right\}
$$

Only the first and last numbers have $f$ values that are multiples of $9 k+r$. Therefore, the middle $2 r \cdot 10^{k}-2$ consecutive positive integers do not have any number $c$ such that $(9 k+r) \mid f(c)$.
(2) When $r=3,6$, $t_{\text {min }}=2(r-1) 10^{k}-1$.
Since $2 r \cdot 10^{k}-1$ consecutive positive integers must include a number $c$ such that $(9 k+r) \mid f(c)$. In this case, $3 \mid f(c)$. And the last three numbers of $F_{1}$ and $F_{2}$ each have only one number that is a multiple of 3, so in the $2 r \cdot 10^{k}$ consecutive numbers $A$, the first and last $2 \times 10^{k}$ numbers can be removed, i.e., in $2(r-1) 10^{k}-1$ consecutive positive integers, there must be a number $c$ such that $(9 k+r) \mid f(c)$.

Below, we construct $2(r-1) 10^{k}-2$ consecutive positive integers such that none of them satisfy $m \mid f(c)$.
Take $a=10^{i}$, satisfying $(9 k+r) \mid(9 i+3)$, then in
$$
a-10^{k}(r-2)+1 \text { to } a+10^{k} r-2
$$

These $2(r-1) 10^{k}-2$ consecutive positive integers do not have any number $c$ such that $(9 k+r) \mid f(c)$.
(3) When $r=9$, $t_{\text {min }}=10^{k+1}-1$.
Similarly to (2), we know that $9 \mid f(c)$ and the last nine numbers of $F_{1}$ and $F_{2}$ each have only one number that is a multiple of 9.

Therefore, in $(2 r-8) 10^{k}-1=10^{k+1}-1$ consecutive positive integers, there must be a number $c$ such that $9(k+1) \mid f(c)$, and in
$$
10^{2 k+1}-10^{k}+1 \text { to } 10^{2 k+1}+9 \times 10^{k}-2
$$

These $10^{k+1}-2$ consecutive positive integers do not have any number $c$ such that $9(k+1) \mid f(c)$.