Is the answer $1010$?

Does anyone have solution

Partition sets in congruence classes $\pmod {17}$. We get $10$ classes with $118$ elements and $7$ classes with $119$ elements so max is $$10 \frac{118}{2}+7 \times 60=1010$$

FISHMJ25 wrote: Partition sets in congruence classes $\pmod {17}$. We get $10$ classes with $118$ elements and $7$ classes with $119$ elements so max is $$10 \frac{118}{2}+7 \times 60=1010$$ We can also print all valid elements according to your solution based on pigeonhole principle

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Partition $2013$ integers into \[\{1,18,35,\ldots ,2007\}, \{2,19,36,\ldots ,2008\}, \{3,19,36,\ldots ,2009\},\ldots ,\{7,24,41,\ldots ,2013\}, \{8,25,42,\ldots ,1997\},\ldots ,\{17,34,51,\ldots ,2006\}\]For the first $7$ sets, there are $119$ elements and for the remaining $10$ sets, there are $118$ elements. We know that by the given condition that we cannot have $2$ consecutive selected elements and given any $4$ consecutive elements, we can select at most $2$ elements which are the $1$st, $3$rd elements or $2$nd, $4$th elements. If we consider the first case for one of the $119$-element sets with each $4$ consecutive elements, we can have at most $60$ selected elements. If we consider the second case similarly, we can have at most $59$ selected elements. Then, implementing the same thing for the $118$-element sets by considering each case, we can have at most $59$ selected elements. Thus, the answer is \[60\times 7+59\times 10=\boxed{1010}.\]