19. If $a_{1}, a_{2}, \cdots, a_{m}$ is an arithmetic sequence with the first term and common difference both positive integers, and each term's decimal representation does not contain the digit 9, find the maximum value of the number of terms $m$.

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19. If $a_{1}, a_{2}, \cdots, a_{m}$ is an arithmetic sequence with the first term and common difference both positive integers, and each term's decimal representation does not contain the digit 9, find the maximum value of the number of terms $m$.
19. Let the decimal representation of $a_{m}$ have $n$ digits. We can add an appropriate number of zeros to the front of each term so that each term has $n$ digits, and we can assume that the leading digits of each term are not all the same (if they are the same, we can remove the leading digit without affecting our proof). Suppose the leading digits of $a_{1}, a_{2}, \cdots, a_{n_{1}}$ are $r_{1}$, the leading digits of $a_{n_{1}+1}, a_{n_{1}+2}, \cdots, a_{n_{2}}$ are $r_{2}$, ..., and the leading digits of $a_{n_{k-1}+1}, a_{n_{k-1}+2}, \cdots, a_{n_{k}}\left(n_{k}=m\right)$ are $r_{k}\left(0=n_{0}<n_{1}<n_{2}<\cdots<n_{k}=m, 0 \leqslant r_{1}<r_{2}<\cdots <r_{k} \leqslant 8\right)$, and let the common difference be $d$. Thus, $a_{n_{i}} \leqslant r_{i} \times 10^{n-1}+\underbrace{88 \cdots 8}_{n-1}=r_{i} \times 10^{n-1}+\frac{8}{9}\left(10^{n-1}-1\right)$, and $a_{n_{i}+1} \geqslant r_{i+1} \times 10^{n-1} \geqslant\left(r_{i}+1\right) \times 10^{n-1}$. Therefore, $d=a_{n_{i}+1}-a_{n_{i}} \geqslant 10^{n-1}-\frac{8}{9}\left(10^{n-1}-1\right)=\frac{1}{9}\left(10^{n-1}-1\right)+1$. Also, $a_{n_{i+1}} \leqslant r_{i+1} \times 10^{n-1}+\frac{8}{9}\left(10^{n-1}-1\right)$, so $a_{n_{i+1}}-a_{n_{i}+1} \leqslant \frac{8}{9}\left(10^{n-1}-1\right)$. Thus, $n_{i+1}-n_{i}=\frac{a_{n_{i+1}}-a_{n_{i}+1}}{d}+1 \leqslant \left[\frac{\frac{8}{9}\left(10^{n-1}-1\right)}{\frac{1}{9}\left(10^{n-1}-1\right)+1}\right]+1=8$. Since the leading digit can only be one of the 9 digits $0,1,2, \cdots, 8$, we have $k \leq 9$. Therefore, $m=n_{k}=\left(n_{k}-n_{k-1}\right)+\left(n_{k-1}-n_{k-2}\right)+\cdots+\left(n_{1}-n_{0}\right) \leqslant \underbrace{8+8+\cdots+8}_{k \uparrow}=8 k \leqslant 72$. On the other hand, the first term is 1, and the common difference is 125. The first 8 terms of this arithmetic sequence are $0001,0126,0251,0376,0501,0626,0751,0876$. After that, every 8 consecutive terms have the leading digit increased by 1, while the other three digits remain unchanged. This way, we can get a total of 72 terms of a positive arithmetic sequence composed of positive integers, and each term does not contain the digit 9. Therefore, the maximum value of $m$ is 72.