If a positive integer has only 0 or 1 as its unit digit in decimal representation, we call this number a "good number". Try to find the smallest positive integer $k$, such that for any positive integer $n$, there is $n=a_{1} \pm a_{2} \pm \ldots \pm a_{k}$, where $a_{i}(1 \leq i \leq k)$ are good numbers.
Three, Solution: $k_{\min }=9$.
First, if $n$ can be expressed as the algebraic sum of $m$ good numbers, then $n$ can also be expressed as the algebraic sum of $m+1$ good numbers. This is because, when $n=a_{1} \pm a_{2} \pm \ldots \pm a_{m}$, we can take a sufficiently large positive integer $r$, such that $10^{r}>\max _{1 \leq 1 \leq m} a_{i}$, and let $a_{m}^{\prime}=a_{m}+10^{r}, a_{m+1}^{\prime}=10^{r}$, then $n=a_{1} \pm a_{2} \pm \ldots \pm a_{m-1} \pm a_{m}^{\prime} \pm 10^{r}$. (*)
First, we prove that $k=9$ satisfies.
For any positive integer $n$, let the digit at the $i$-th position be $k$, then let the digit at the $i$-th position of $a_{1}, a_{2}, \ldots, a_{k}$ be 1, and the digit at the $i$-th position of $a_{k+1}, \ldots, a_{9}$ be 0 (if $k=0$, then the digit at the $i$-th position of $a_{1}, a_{2}, \ldots, a_{9}$ is 0; if $k=9$, then the digit at the $i$-th position of $a_{1}, a_{2}, \ldots, a_{9}$ is 1). Then $n=a_{1}+a_{2}+\ldots+a_{9}$ (if there exists $1 \leq j \leq 9$ such that $a_{j}=0$, then remove this number, by (*) we can express $n$ as the sum of 9 good numbers), hence $k=9$ holds;
Next, we prove that $k \leq 8$ does not hold. Take $n=10203040506070809$, if $n$ can be expressed as the algebraic sum of 8 good numbers, let $n=a_{1}+a_{2}+\ldots+a_{j}-\left(a_{j+1}+a_{j+2}+\ldots+a_{8}\right)$, we set $b_{1}=a_{1}+a_{2}+\ldots+a_{j}$, $b_{2}=a_{j+1}+a_{j+2}+\ldots+a_{8}$, then $b_{1}=n+b_{2}$. Since the digits of $b_{1}$, $b_{2}$ do not exceed $j, 8-j$, and $n$ has a digit that must be $j+1$, denoted as the $m$-th digit, hence in the calculation of $n+b_{2}$, the $m$-th digit must carry over, thus the digit of $b_{2}$ at the $m$-th position is not less than $9-j$ (since the $(m+1)$-th digit of $n$ is 0, hence the $(m+1)$-th digit does not carry over in the calculation of $n+b_{2}$), which contradicts that the digits of $b_{2}$ do not exceed $8-j$.
In summary, $k_{\min }=9$.