Let's find 8 positive integers less than 100 such that among the 255 sums that can be formed from them, there are no equal sums. (We include the 8 numbers themselves as one-term sums.)
I. solution. We call a sequence of positive integers "economical" if it has the property described in the problem, that is, there are no equal sums among the sums that can be formed from its elements. For example, with a set of weights of such a sequence, we cannot measure equal masses if we place different groups of weights on the scale pan. The sequence of powers of 2, for example, is clearly economical, since any of its elements is greater than the sum of the smaller ones, but this is precisely the reason why this sequence grows too quickly: its eighth element, $2^{7}$, is already greater than 100. However, the powers of 2 can be used to create an economical sequence from numbers less than 100.

The first six non-negative powers of 2, the sequence $1,2,4,8,16,32$ is thus economical, and the sums that can be formed from its elements are precisely the integers between 1 and $2^{6}-1=63$. Therefore, the sums that can be formed from the multiples of 3, the numbers 3, 6, 12, 24, 48, 96, do not contain equal sums, and these sums are precisely the numbers divisible by 3 between 3 and $3 \cdot 63=189$.

We claim that if we add the numbers 95 and 97 to these six numbers, the resulting 8-element sequence will still be economical. Indeed, in addition to the existing ones, we can get the following new sums:

1. 97 is a term in them, but 95 is not; these sums give a remainder of 1 when divided by 3, and any two of them are different.
2. 95 is a term in them, but 97 is not; these sums give a remainder of 2 when divided by 3, and there are no equal ones among them.
3. Both new numbers, 95 and 97, are terms in them. These sums are divisible by 3, but not less than $95+97=192$; of course, these sums are also different from each other.

It can be seen that the new sums created by adding the two new numbers are different from the previous ones and from each other, and thus the numbers $3,6,12,24,48,95,96,97$ indeed provide a solution to the problem.

Remark. It is generally true that the sequence $3,3 \cdot 2,3 \cdot 2^{2}, \ldots, 3 \cdot 2^{k-1}, 3 \cdot 2^{k}-1,3 \cdot 2^{k}, 3 \cdot 2^{k}+1$ is economical. The following solution will show that this is not the best in the sense that the largest element of the sequence can be reduced. Below we give a sufficient condition for an 8-element sequence to be economical - the condition can generally be formulated - and then use it to construct a suitable sequence.

II. solution. It seems natural to try to prescribe the magnitude of the sums that can be formed: the conditions derived from this are expected to help in the construction of the sequence.

Let us try, then, to arrange the sums that can be formed as follows:

a) if two sums consist of a different number of terms, then the one with fewer terms should be smaller;

b) if two sums consist of the same number of terms, then the one with the smallest term - after the possible common terms are omitted - should be smaller.

If there is such a sequence of numbers, then it is clearly economical, only it is not clear whether the two conditions can be met simultaneously. Note that both conditions somehow ensure that the elements of the sequence are not too large relative to each other - for example, the largest element is smaller than the sum of the two smallest ones - and it is expected that through these, we will be able to find a solution to the problem consisting of numbers less than 100.

Let the elements of the sequence be $a_{1}, a_{2}, \ldots, a_{s}$ in increasing order. Due to the arrangement of the elements, the second condition is satisfied if

$$
\begin{aligned}
a_{i}+a_{8} & < a_{7}+a_{6}+a_{5}+a_{4}+a_{3}+a_{2}+a_{1} \quad \text{for} \quad i=1,2,3,4,5,6,7.
\end{aligned}
$$

Starting from the numbers $a_{8}>a_{7}>a_{6}>a_{5}>a_{4}>a_{3}>a_{2}>a_{1}$, it is clear that $a_{6}>a_{5}>a_{4}>a_{3}>a_{2}>a_{1}$, and on the other hand, the sums of the same number of terms will indeed be different.

We still need to ensure that the elements of the sequence are positive numbers, and of course, the condition regarding the arrangement of sums with different numbers of terms. This latter surprisingly follows from the seemingly less assertive

$$
a_{1}+a_{2}+a_{3}+a_{4}>a_{6}+a_{7}+a_{8}
$$

condition; it is sufficient to ensure that the smallest four-term sum is greater than the largest three-term sum.

Indeed, due to the arrangement of the elements,

$$
a_{2}-a_{8}<a_{1} \quad \text{and} \quad a_{3}-a_{8}<a_{1} \quad \text{and} \quad a_{4}-a_{8}<a_{1} \quad \text{and} \quad a_{5}-a_{8}<a_{1} \quad \text{and} \quad a_{6}-a_{8}<a_{1} \quad \text{and} \quad a_{7}-a_{8}<a_{1}.
$$

Therefore, for example, starting from $a_{8}=88$, we get the sequence

$88,87,86,84,81,75,64,42$

; this sequence, according to the proven properties of the construction, meets the requirements of the problem.

Remarks. 1. If we slightly modify the conditions defining the sequence (it is sufficient to calculate $a_{1}$ from the relation $a_{1}=a_{2}+a_{3}+a_{4}-a_{7}-a_{8}-1$ instead of (4), for if the sum of the eight numbers is odd, then we clearly cannot get equal four-term sums; on the other hand - and this the reader should think about - if instead of (5) we only write that the smallest four-term sum should be 1 less than the largest three-term sum, then there will still be no equal sums among the sums with different numbers of terms), then we get the sequence 84, 83, 82, 80, 77, 71, 60, 40.

We could not find an 8-element economical sequence with the largest element less than 84 using a calculator, although the program did not examine all possibilities.

In problem Sz. 55, a program had to be written that generates an economical sequence from two-digit numbers. The vast majority of the submitted programs are not suitable for solving the problem, as they would run for years in the best case. The authors did not seem to be bothered by this fact.

2. The problem is closely related to problem F. 2584 (see the solution in the 304th page of the 7th issue of 1986). There we proved that an economical sequence of 10 elements consisting of two-digit numbers does not exist. The proof there does not work for a 9-element sequence, on the other hand, none of the current constructions provide a 9-element economical sequence.