1. Let $S_n=\sum_{r=1}^n \tan^{-1}\left(\frac{1}{1+r+r^2}\right).$ We need to find $\lim_{n\to\infty} \tan(S_n).$

2. Simplify the term inside the sum using the identity $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$ when the angles lie in the principal range.

Take $a=r+1,\qquad b=r.$ Then $\tan^{-1}(r+1)-\tan^{-1}(r)=\tan^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}\right)=\tan^{-1}\left(\frac{1}{1+r+r^2}\right).$ So each term becomes $\tan^{-1}\left(\frac{1}{1+r+r^2}\right)=\tan^{-1}(r+1)-\tan^{-1}(r).$

3. Therefore the sum telescopes: $S_n=\sum_{r=1}^n \big(\tan^{-1}(r+1)-\tan^{-1}(r)\big).$ Hence $S_n=\tan^{-1}(n+1)-\tan^{-1}(1).$

4. Now compute $\tan(S_n)$: $\tan(S_n)=\tan\big(\tan^{-1}(n+1)-\tan^{-1}(1)\big).$ Using $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},$ we get $\tan(S_n)=\frac{(n+1)-1}{1+(n+1)(1)}=\frac{n}{n+2}.$

5. Take the limit: $\lim_{n\to\infty} \tan(S_n)=\lim_{n\to\infty} \frac{n}{n+2}=1.$

Therefore, the required integer is $\boxed{1}.$