The 64th question: Find all real numbers $\mathrm{a}(\mathrm{a} \geq 1)$, such that for every sequence of pairwise distinct positive integers $\left\{a_{1}, a_{2}, \ldots\right\}$, if for any positive integer $\mathrm{n}$, we have $\mathrm{a}_{\mathrm{n}} \leq \mathrm{an}$, then the sequence contains infinitely many terms whose sum of digits in base 4038 is not divisible by 2019.
Question 64:
Solution Method One: For any natural number $b$, let $S(b)$ be the sum of the digits of $b$ in base 4038. If $2019 \mid S(b)$, then $b$ is called a good number; if $2019 \nmid S(b)$, then $b$ is called a bad number.
(i) First, prove that when $\mathrm{a}\frac{4038^{\mathrm{k}}-1}{\mathrm{a}}-1$ terms are no greater than $4038^{\mathrm{k}}-1$, i.e., in the interval $\left[1,4038^{\mathrm{k}}-1\right]$, at least $\frac{4038^{\mathrm{k}}-1}{\mathrm{a}}-1$ terms fall into the sequence. Among these terms, at least
$$
\frac{4038^{k}-1}{a}-1-\left(2 \cdot 4038^{k-1}-1\right)=\frac{2(2019-a) 4038^{k-1}}{a}
$$

are bad numbers.
When $k \rightarrow+\infty$, $\frac{2(2019-\mathrm{a}) 4038^{k-1}}{a} \rightarrow+\infty$, hence the sequence has infinitely many bad numbers.
(ii) Next, prove that when $a \geq 2019$, there exists a sequence $\left\{a_{1}, a_{2}, \ldots\right\}$ that satisfies the conditions, and only a finite number of terms in the sequence are bad numbers.
In fact, for any $\mathrm{i} \in\{0,1,2, \ldots, 2019\}$, and any natural number $\mathrm{k}$,
$$
S(4038 k+i), S(4038 k+i+1), \cdots, S(4038 k+i+2018)
$$

are 2019 consecutive positive integers, forming a complete residue system modulo 2019.
First, take $\mathrm{a}_{1}=1$. Then, for any positive integer $\mathrm{m}$, take
$\mathrm{a}_{2 \mathrm{~m}} \in\{4038(\mathrm{~m}-1)+2019,4038(\mathrm{~m}-1)+4037\}$
$\mathrm{a}_{2 \mathrm{~m}+1} \in\{4038 \mathrm{~m}, 4038 \mathrm{~m}+2018\}$ and ensure that $2019\left|S\left(a_{2 m}\right), 2019\right| S\left(a_{2 m+1}\right)$.
Clearly, the sequence $\left\{a_{1}, a_{2}, \ldots\right\}$ constructed in this way satisfies the given conditions, and except for $a_{1}$, all other numbers are good numbers.
In summary, the required $\mathrm{a}$ is any real number in $[1,2019)$.
Solution Method Two: For any natural number $b$, let $S(b)$ be the sum of the digits of $b$ in base 4038. If $2019 \mid S(b)$, then $b$ is called a good number; if $2019 \nmid S(b)$, then $b$ is called a bad number.

Notice that for any natural number $n, S(2019 n), S(2019 n+1), \cdots, S(2019 n+2018)$ are 2019 consecutive positive integers, and exactly one of them is divisible by 2019, i.e., there is exactly one good number, denoted as $b_{n}$. Clearly, $b_{0}=0$; and for any positive integer $n, 2019 n \leq b_{n}\frac{2019(\mathrm{n}+1)}{\mathrm{a}}-1$ terms are no greater than $2019(\mathrm{n}+1)$, i.e., in the interval $[1,2019(n+1))$, at least $\frac{2019(n+1)}{a}-1$ terms fall into the sequence. Among these terms, at least
$$
\left(\frac{2019(n+1)}{a}-1\right)-n=\frac{(2019-a)(n+1)}{a}
$$

are bad numbers.
When $n \rightarrow \infty$, $\frac{(2019-a)(n+1)}{a} \rightarrow+\infty$, hence the sequence $\left\{a_{1}, a_{2}, \ldots\right\}$ has infinitely many bad numbers.
(ii) Next, prove that when $a \geq 2019$, there exists a sequence $\left\{a_{1}, a_{2}, \ldots\right\}$ that satisfies the conditions, and only a finite number of terms in the sequence are bad numbers.

In fact, first take $a_{1}=1$, then take $a_{n}=b_{n-1}(n \geq 2)$. Then for any positive integer $n, a_{n}=b_{n-1}$ $<2019 n \leq$ an. And except for $a_{1}$, all other $a_{n}$ are good numbers.
In summary, the required $\mathrm{a}$ is any real number in $[1,2019)$.