The sum of all $64$ numbers is their mean times the number of numbers, which is $64 \times 64 = 4096$. Similarly the sum of the first $36$ numbers is $36^2 = 1296$. Therefore the sum of the last $28$ numbers is $64^2-36^2= 4096 - 1296 = 2800$. Therefore the mean of the last $28$ numbers is $\frac{2800}{28}=100$.

*This problem is taken from the UKMT Mathematical Challenges.*