Analysis and Solution: Find a sufficient condition to ensure that the selected boxes can accommodate all the documents.
Consider all documents with numbers $a$, $b$, and $c$ (where $a$, $b$, and $c$ may be the same) as their codes. To accommodate these documents, there must be a box with at least two of these numbers (for example, $a$ and $b$) as its code. To ensure that documents with different permutations of the numbers can be accommodated, a sufficient condition is that boxes with codes $\overline{a b}$ and $\overline{b a}$ are both selected. This means that "any two numbers among any three numbers are selected, and all permutations of these two numbers are selected." This leads to the pigeonhole principle: 3 numbers are divided into 2 sets, so one set must contain two numbers. Therefore, divide $0,1,2, \cdots, 9$ into two subsets $A$ and $B$, and ensure that any 2-number combination within the same subset is selected, and all permutations (i.e., all 2-element permutations with repetition) of these 2-number combinations are selected (to ensure that any order can be accommodated), then these codes meet the requirement.

Let $|A|=k$ and $|B|=10-k$. The number of 2-element permutations with repetition in $A$ and $B$ are $k^{2}$ and $(10-k)^{2}$, respectively. Therefore, the total number of such 2-element permutations is $k^{2}+(10-k)^{2}=2(k-5)^{2}+50 \geqslant 50$.

Taking $k=5$, i.e., $|A|=|B|=5$, for example, $A=\{0,1,2,3,4\}$ and $B=\{5,6,7,8,9\}$. Then, the 2-element permutations with repetition in $A$ and $B$ each have 25, making a total of 50 codes that meet the requirement.

Next, we prove that $k \geqslant 50$.
Suppose that among the $k$ selected boxes, the number of boxes with 9 as the first digit is the least, say $m$ (parameter) boxes, denoted as $\overline{9 a_{i}}$ $(i=1,2, \cdots, m)$. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$. Take any two numbers $a$ and $b$ not in $A$, and consider the document with code $\overline{9 a b}$. Since $a$ is not in $A$, there is no box with code $\overline{9 a}$. Similarly, there is no box with code $\overline{9 b}$. Therefore, the box with code $\overline{a b}$ must be selected. Note that $a$ and $b$ each have $10-m$ possible values, so there should be $(10-m)^{2}$ such boxes, and none of these boxes have $a_{1}, a_{2}, \cdots, a_{m}$ as the first digit. By the "minimum property," there are at least $m$ boxes with $a_{1}, a_{2}, \cdots, a_{m}$ as the first digit, so there are at least $m^{2}$ such boxes. Therefore, $k \geqslant m^{2}+(10-m)^{2} \geqslant \frac{1}{2}[m+(10-m)]^{2}=50$.

In conclusion, the minimum value of $k$ is 50.
Example 4 There are 1000 certificates numbered $000,001, \cdots, 999$ and 100 boxes numbered 00, $01, \cdots, 99$. If the number of a box can be obtained by removing one digit from the number of a certificate, then the certificate can be placed in that box. If $k$ boxes can be chosen to accommodate all certificates, find the minimum value of $k$.