jmc

Let $\omega$ be a root of $x^2+x+1=0,$ so ${\omega }^2+\omega +1=0.$ Then by the factor theorem, $x^{2n}+1+(x+1{)}^{2n}$ is divisible by $x^2+x+1$ if and only if ${\omega }^{2n}+1+(\omega +1{)}^{2n}=0.$

Since $\omega +1=-{\omega }^2,$ $${\omega }^{2n}+1+(\omega +1{)}^{2n}={\omega }^{2n}+1+(-{\omega }^2{)}^{2n}={\omega }^{4n}+{\omega }^{2n}+1.$$From the equation ${\omega }^2+\omega +1=0,$ $(\omega -1)({\omega }^2+\omega +1)={\omega }^3-1,$ so ${\omega }^3=1.$

We divide into the cases where $n$ is of the form $3k,$ $3k+1,$ and $3k+2.$

If $n=3k,$ then $$\begin{array}{rl}{\omega }^{4n}+{\omega }^{2n}+1& ={\omega }^{12k}+{\omega }^{6k}+1\\ & =({\omega }^3{)}^{4k}+({\omega }^3{)}^{2k}+1\\ & =1+1+1=3.\end{array}$$If $n=3k+1,$ then $$\begin{array}{rl}{\omega }^{4n}+{\omega }^{2n}+1& ={\omega }^{12k+4}+{\omega }^{6k+2}+1\\ & =({\omega }^3{)}^{4k+1}\omega +({\omega }^3{)}^{2k}{\omega }^2+1\\ & =\omega +{\omega }^2+1=0.\end{array}$$If $n=3k+2,$ then $$\begin{array}{rl}{\omega }^{4n}+{\omega }^{2n}+1& ={\omega }^{12k+8}+{\omega }^{6k+4}+1\\ & =({\omega }^3{)}^{4k+2}{\omega }^2+({\omega }^3{)}^{2k+1}\omega +1\\ & ={\omega }^2+\omega +1=0.\end{array}$$Hence, $x^{2n}+1+(x+1{)}^{2n}$ is divisible by $x^2+x+1$ if and only if $n$ is of the form $3k+1$ or $3k+2,$ i.e. is not divisible by 3. In the interval $1\le n\le 100,$ there are $100-33=\boxed{67}$ such numbers.