Let $2 S$ be the total weight of some set of weights. We call a natural number $k$ average if we can choose $k$ weights from the set such that their total weight is $S$. What is the maximum number of average numbers that a set of 100 weights can have?
Note that if the number $m$ is average, then the number $100-m$ is also average. Therefore, if the number 1 is not average, then the number 99 is also not average, and the number of average numbers is no more than 97 (the number 100 is also not average). If, however, the number 1 is average, then the weight of one of the weights is $S$ and, consequently, only 99 is also an average number. Thus, the number of average numbers does not exceed 97. Let's provide an example of a set of 100 weights with weights $a_{1}, \ldots, a_{100}$, for which all numbers from 2 to 98 (a total of 97 numbers) are average.

Let $a_{1}=a_{2}=1, a_{n+2}=a_{n}+a_{n+1}, n=1,2, \ldots, 97,$ - consecutive Fibonacci numbers and $S=a_{1}+a_{2}+\ldots+a_{98}$. Choose $a_{100}=S-a_{99}$. Then the total weight of all weights is $2 S$ and, at the same time,

$a_{100}+a_{99}=a_{100}+a_{98}+a_{97}=a_{100}+a_{98}+a_{96}+a_{95}=a_{100}+a_{98}+a_{96}+a_{94}+a_{93}=\ldots=a_{100}+a_{98}+a_{96}+a_{94}+$ $a_{92}+a_{90}+\ldots+a_{6}+a_{4}+a_{2}+a_{1}=S$

Therefore, the numbers $2,3,4, \ldots, 51$ are average. But then the numbers $100-2=98,100-3=97, \ldots, 100-48=52$, that is, all numbers from 2 to 98 are average.

## Answer

97.

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