4. A magician has one hundred cards, each with a number from 1 to 100. He places these one hundred cards into three boxes, one red, one white, and one blue. Each box must contain at least one card.

A participant selects two of the three boxes and then picks one card from each of the selected boxes, announcing the sum of the numbers on the two cards. Knowing this sum, the magician can identify which box was not selected.

How many ways are there to place the cards so that the trick always works? (Two methods are considered different if at least one card is placed in a different colored box.)


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Solution: There are 12 different methods. Consider the integers from 1 to 100. For simplicity, define the color of the box in which integer $i$ is placed as the color of that integer. Use $\mathrm{r}$ to represent red, $\mathrm{w}$ to represent white, and $\mathrm{b}$ to represent blue.
(1) There exists some $i$ such that $i, i+1, i-2$ have mutually different colors, for example, $\mathrm{wb}$. Since $i+(i+3)=(i+1)+(i+2)$, the color of $i \div 3$ cannot be the color of $i+1$, which is $\mathrm{w}$, nor can it be the color of $i+2$, which is $\mathrm{w}$. It can only be $\mathrm{r}$. Therefore, as long as three adjacent numbers have mutually different colors, the color of the next number can be determined. Furthermore, this pattern of three numbers' colors must repeat: after rwb, it must be $r$, then $w, b, \cdots$ and so on. Similarly, this process also holds for the opposite direction: before rwb, it must be $b, \cdots$ and so on.

Therefore, it is only necessary to determine the colors of 1, 2, and 3. There are 6 different ways to do this, and all 6 methods can make the magic trick successful because their sums $r+w, w+b, b+r$ give mutually different remainders modulo 3.
(2) There do not exist three consecutive numbers with mutually different colors. Assume 1 is red. Let $i$ be the smallest number that is not red, and assume $i$ is white. Let $k$ be the smallest blue number. By assumption, $i+11$ must be red, and from $t+99=(t-1)+100$, it follows that $t-1$ is blue, which contradicts the assumption that only 100 is blue.

Therefore, the colors of these numbers must be rww…wwb. This method indeed works:

If the sum of the numbers on the two selected cards $\leqslant 100$, the box from which no card was selected must be blue;

If the sum of the numbers is $101$, the box from which no card was selected must be white;

If the sum of the numbers $>101$, the box from which no card was selected must be red.
Finally, there are 6 different ways to arrange the colors in the above manner. Therefore, the answer is 12.