The arithmetic mean of a sequence of natural numbers beginning with one is $ \frac {n + 1}{2}$. For $ n = 99\rightarrow101$, the mean is very close to the mean with one missing number. This suggests that the mean of the initial numbers is either $ 50, 50.5, 51$. Arithmetic mean does not vary if we shorten the sequence by eliminating the front and end numbers in pairs. Now we need to find how short to cut it. $ 50.55$ suggests that there are $ 21$ initial terms because of the $ \frac {11}{20}$ found in the mean without one term. This eliminates numbers with a mean of $ 50.5$ since that occurs from an even number of natural numbers. Furthermore, the greatest difference that can result from removing one integer from a sequence is $ .5$, which eliminates $ 50$ as a possible mean. Thus, the mean must be $ 51$. Removing the largest number from the sequence ($ 61$) results in a shift of the mean to $ 50.5$. Since we want a number $ \frac {1}{20}$ away from this mean, we can remove the second largest number. The initial set of numbers was $ 41\rightarrow61$ and the removed number was $ 60$.