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What is the maximum amount of numbers that can we choose from the set $\{1,2,3,\dots ,50\}$ that there isn't any two numbers that their sum is a multiply of $7$.

I take the number $7$ and also the numbers that $a\equiv 1,2,3\pmod 7$ and I got $23$ numbers.Isn't there a bigger case?

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    No there isn't, but there are other cases that are equally large. – 2017-02-16
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    @ErickWong Then my answer is true? – 2017-02-16
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    It is the correct value, but you ought to think about how to *prove* there is nothing larger, as that will eliminate your doubt. – 2017-02-16

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If you take $24$ numbers, at least $23$ of them are in $A=\{1,2,\ldots,49\}$. Let $A_k=\{7k+1,7k+2,\ldots,7k+7\}$ for $k=0,\ldots,6$. At least two of the $A_k$'s has four of the chosen numbers, or at least one of the $A_k$'s has five.

Anyway, consider the pairs $(7k+1,7k+6)$, $(7k+2,7k+5)$ and $(7k+3,7k+4)$. This pairs cover the first six elements of $A_k$. The latter case implies that two of the chosen numbers are in the same pair, and hence their sum is a multiple of $7$.

Similarly, the former case implies that in one of the $A_k$ we have chosen two numbers of the same pair, or we have chosen the last number (the one that is multiple of $7$) in both sets.