Let the sum of the remaining numbers be $ A$. Then, because there are $ n - 4$ numbers remaining after $ 4$ numbers are removed, have $ \frac {A}{n - 4} = 51.5625 = \frac {825}{16}$ Now, the sum of the $ 4$ even numbers removed is greater than $ 10 = 1 + 2 + 3 + 4$ but less than $ n + (n - 1) + (n - 2) + (n - 3)$ because the numbers removed are all even and belong to the set $ \{1,2,...,n\}$. The sum of all the numbers in set $ \{1,2,...,n\}$ is $ \frac {n(n + 1)}{2}$. Then, $ \frac {n(n + 1)}{2} - (n + (n - 1) + (n - 2) + (n - 3)) < A < \frac {n(n + 1)}{2} - 10$ Simplifying, get: $ \frac {(n - 3)(n - 4)}{2} < A < \frac {(n - 4)(n + 5)}{2}$ Then: $ \frac {n - 3}{2} < \frac {A}{n - 4} < \frac {n + 5}{2}$ But $ \frac {A}{n - 4} = \frac {825}{16}$ so: $ \frac {n - 3}{2} < \frac {825}{16} < \frac {n + 5}{2}$ Clearing denominators, we get $ 98.125 < n < 106.125$, and as $ n$ is a positive integer, $ 99 < = n < = 106$. On the other hand, if the four consecutive numbers are $ 2(k - 1),2(k),2(k + 1),2(k + 2)$($ k$ is an integer) their sum is $ 2(4k + 2)$ the sum of all the numbers from $ 1$ to $ n$ is: $ A + 2(4k + 2) = \frac {n(n + 1)}{2}$ But $ A = \frac {825}{16}(n - 4)$, so: $ \frac {825}{16}(n - 4) + 2(4k + 2) = \frac {n(n + 1)}{2}$ (*) Because $ 2(4k + 2)$, $ \frac {n(n + 1)}{2}$ are integers, $ \frac {825}{16}(n - 4)$ is an integer, so $ 16|(n - 4)$ so $ n\equiv 4\pmod {16}$. Then, the only such value of $ n$ such that $ 99 \le n \le 106$ is $ n = 100$. Then, substituting $ n = 100$ into (*) have: $ 2(4k + 2) = 100$ then $ k = 12$ and so the four consecutive even integers removed are $ 22,24,26,28$.