South African Mathematics Olympiad Second Round

The first $50$ even positive integers are $2,4,6,\dots ,96,98,100$, and their sum is $50\times \frac{1}{2}(2+100)=50\times 51=2550$. To reduce the sum to $2016$ we must subtract $534$. The average of the numbers removed must be less than $100$, so we need to remove at least six. To use as few numbers as possible, we first remove the five largest numbers from $100$ down. Now $100+98+96+94+92=480$, which is nearly there, so to bring the total subtracted to $534$ we finally need to remove $54$.

[There are, of course, many other combinations of five integers with a sum of $534$ that can be removed. ]