7. The lye trader Matti wrote several three-digit palindromic numbers on the board. It turned out that no two of them add up to a palindrome. Could he have written more than half of all three-digit palindromes? Recall that a palindrome is a number that remains the same when its digits are reversed.
Solution. No. There are $9 \cdot 10$ three-digit palindromes in total (9 options for the first digit, 10 for the second, and the third is uniquely determined by the first). We will prove that at least 45 palindromes are missing from the board.

If $a, b, c, d$ are such digits that $a+c=b+d=11$, then

$$
\overline{a b a}+\overline{c d c}=1221
$$

which means all palindromes consisting of digits from 2 to 9 can be paired, and from each pair, no more than one can be written. There are 64 such palindromes in total (8 options for the first digit and 8 for the second), so at least 32 of them are missing.

Now consider the following group of 13 palindromes:

$$
\begin{gathered}
\text { 101, 111, 121, 131, 141, 151 } \\
202,303,404,505 \\
212,313,414
\end{gathered}
$$

No more than one of them can be written, as adding any two of them does not result in a carry, so the sum will be a palindrome. Therefore, at least 12 of them are missing.

Finally, there are also pairs $(161,616),(171,717)$, and $(181,818)$, and from each pair, at least one palindrome is missing. So, at least 3 in total.

In the end, Vasya will be missing at least $32+12+3=47$ palindromes, which is more than 45.

Moreover, this estimate seems quite rough, so there may be many unpleasant solutions that need to be listened to with extreme caution, requiring the presenter to formulate strict statements and clearly explain which cases they are considering, etc. And generally, it is not advisable to hesitate to seek help if it is unclear whether it is a fake or not.