4. The number 100 is represented as the sum of several two-digit numbers, and in each addend, the digits are swapped. What is the largest number that could result from the new sum?
Solution. If the digits in the two-digit number $\overline{b a}$ are swapped, the number increases by $9(a-b)$. Therefore, the new sum is $S=100+9 U-9 D$, where $D$ is the sum of the tens digits, and $U$ is the sum of the units digits in the original addends. Since $10 D+U=100$, then $S=1000-99 D$. Thus, we need to minimize $D$ - the sum of the tens digits. It can be 6, for example, $100=19 \cdot 4+24$, but not less. Suppose $D<6$, then there are no more than 5 numbers. Thus, $U$ is no more than $5 \cdot 9=45$, and the sum of the numbers before swapping the digits does not exceed $10 \cdot 5+45=95<100$. Contradiction. Therefore, the minimum possible $D$ is 6, and the maximum sum is $1000-99 \cdot 6=406$.