Hint: $E=\frac{{{n}^{2021}}+101}{{{n}^{2}}+n+1}=\frac{{{n}^{2021}}-{{n}^{2}}+{{n}^{2}}+101}{{{n}^{2}}+n+1}=\frac{{{n}^{2}}({{n}^{2019}}-1)+{{n}^{2}}+101}{{{n}^{2}}+n+1}=\text{integer}+\frac{{{n}^{2}}+n+1+100-n}{{{n}^{2}}+n+1}$ So $E=\text{integer}+1-\frac{n-100}{{{n}^{2}}+n+1}\in \mathbb{Z}\Rightarrow \frac{n-100}{{{n}^{2}}+n+1}\in \mathbb{Z}$

note that \[(n^3-1)=(n-1)(n^2+n+1)\]with this we get that $n^2+n+1$ id divisor of $n^3-1$. Because $(a^k-b^k)$ is divisor of $(a^{3k}-b^{3k})$ for $k$ positive integer, hence $n^3-1$ is divisor of $(n^3)^{673}-1=n^{2019}-1$. Because \[\frac{n^{2021}+101}{n^2+n+1}=\frac{n^2(n^{2019}-1)+n^2+101}{n^2+n+1}\]Hence \[\frac{n^2+101}{n^2+n+1}\]is an integer, and thus \[x=\frac{n-100}{n^2+n+1}\]is an integer, just note when $n>9$ the number is $-1<x<0<x<1$. So its only when $n-100=0$, $n=100$ and just check for $0<n\leq9$ The solution for $n=1,2,9,100$