jmc

Let $b_n=a_{n+1}-a_n.$ Then $$\begin{array}{rl}{b}_n& =(11a_n-(n+1))-a_n\\ & =10a_n-(n+1)\\ & =10(11a_{n-1}-n)-(n+1)\\ & =11(10a_{n-1}-n)-1\\ & =11b_{n-1}-1.\end{array}$$Hence, $$b_n-\frac{1}{10}=11b_{n-1}-\frac{11}{10}=11\left(b_{n-1}-\frac{1}{10}\right).$$If $b_1<\frac{1}{10},$ then the sequence $b_1,$ $b_2,$ $\dots$ is decreasing and goes to $-\infty ,$ so the sequence $a_1,$ $a_2,$ $\dots$ goes to $-\infty$ as well.

Hence, $b_1\ge \frac{1}{10}.$ Then $a_2-a_1\ge \frac{1}{10},$ so $$11a_1-2=a_2\ge a_1+\frac{1}{10}.$$This implies $a_1\ge \frac{21}{100}.$

If $a_1=\frac{21}{100},$ then the sequence $a_1,$ $a_2,$ $\dots$ is increasing (since $b_n=\frac{1}{10}$ for all $n$), so all the terms are positive. Therefore, the smallest possible value of $a_1$ is $\boxed{\frac{21}{100}}.$