Let the even numbers excluded be $2k,2k+2,2k+4$ and $2k+6$. Then we know that

${\displaystyle \sum _{i=1}^{n}i-(8k+12)=51.5625\times (n-4)\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1).}$

Hence:

${\displaystyle [\sum _{i=5}^{n}i-(8k+12)]/(n-4)\ge 51.5625\ge [\sum _{i=1}^{n-4}i]/(n-4).}$

It follows that:

${\displaystyle n^2-102.125n+392.5\ge 0\hspace{0.5em}\mathrm{quad}(2)}$

and

${\displaystyle 51.5625\ge \frac{1}{2}(n-3)\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(3).}$

Solving the quadratic equation $n^2-102.125n+392.5=0$ gives

${\displaystyle n=\frac{102.125\pm 94.125}{2}=4\hspace{0.5em}\mathrm{or}\hspace{0.5em}98.125}$

From (2) and (3) we have $98\le n\le 106.$ From (1):

${\displaystyle n^2-102.125n-16k+388.5=0}$

and so

${\displaystyle k=\frac{n^2-102.125n+388.5}{16}.}$

Using a spreadsheet to calculate $k$ for $n\in [98,106]$ we find that the only integer value of $k$ is $k=11$ corresponding to $n=100$. The unique solution to the problem is $n=100$ and the even numbers excluded are 22,24,26 and 28.