4. On the table, there are 100 weights of different masses. A weight is called successful if its mass is equal to the sum of the masses of some two other weights on the table. What is the smallest number of successful weights that would allow us to assert with certainty that the masses of some two weights differ by at least a factor of three?
Answer: 87. Let there be 87 lucky numbers. If one of the two addends of a lucky number $a=b+c$ is itself lucky, for example, $b=d+e$, then $a=c+d+e \geqslant 3 \min (c, d, e)$, which means there will be two numbers differing by at least a factor of three. If, however, all the addends of the eighty-seven lucky numbers are among the 13 unlucky ones, then each lucky number is one of the $13 \cdot 12 / 2=78$ pairwise sums of these - a contradiction.

Example with 86 lucky numbers: take 14 almost identical numbers and 86 of their $14 \cdot 13 / 2=91$ pairwise sums and declare them to be lucky numbers. For instance, consider the numbers $1+\varepsilon, 1+2 \varepsilon, 1+2^{2} \varepsilon, \ldots, 1+2^{12} \varepsilon$. All their pairwise sums are distinct, greater than the original numbers, and the largest of them $2+2^{12} \varepsilon+2^{13} \varepsilon=2+3 \cdot 2^{12} \varepsilon < 3(1+\varepsilon)=3+3 \varepsilon$ for a very small $\varepsilon$.