Ends in 00

Let the property $P$ be the property that a set has a pair with sum or difference which is a multiple of $100$ .

Consider any set of elements mod $100$ ; note that whether any set has the desired property will remain unchanged when we take all elements mod $100$ .

For a set without $P$ , then for any element $r$ there cannot be another element equal to $r$ or $100-r$ . Hence there are no repeats.

Firstly, for any $n \leq 51$ the set $\{0,1,2,3,4,\ldots, n-1\}$ does not satisfy $P$ .

For $n = 52$ , if we assume there is a set without property $P$ , then we split $\{0,1,2,3,4,\ldots,99\}$ into $51$ sets of $\{0\}, \{1,99\}, \ldots \{r,100-r\}, \ldots, \{49,51\}, \{50\}$ .

By the pigeonhole principle, if we choose $52$ elements, there must exist $2$ elements which belong to the same pair even if there are no repeats. These $2$ elements make the set have property $P$ , hence we have proven that all $52$ element sets have property $P$ .