Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.

Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.
Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5, 14, or 23. We check $s(2008-5)=s(2003)=5, s(2008-14)=s(1994)=23, s(2008-23)=s(1985)=$ 23. Thus $a_{n-1}$ can equal 1985 or 2003. As above $2 a_{n-2} \equiv a_{n-1} \equiv 5 \equiv 14$, so $a_{n-2} \equiv 7$. But $a_{n-2}<2003$, so $s\left(a_{n-2}\right) \leq 28$ and thus $s\left(a_{n-2}\right)$ can equal 16 or 25. Checking as above we see that the only possibility is $s(2003-25)=s(1978)=25$. Thus $a_{n-2}$ can be only 1978. Now $2 a_{n-3} \equiv a_{n-2} \equiv 7 \equiv 16$ and $a_{n-3} \equiv 8$. But $s\left(a_{n-3}\right) \leq 27$ and thus $s\left(a_{n-3}\right)$ can equal 17 or 26. The check works only for $s(1978-17)=s(1961)=17$. Thus $a_{n-3}=1961$ and similarly $a_{n-4}=1939 \equiv 4, a_{n-5}=1919 \equiv 2$ (if they exist). The search for $a_{n-6}$ requires a residue of 1. But $a_{n-6}<1919$, so $s\left(a_{n-6}\right) \leq 27$ and thus $s\left(a_{n-6}\right)$ can be equal only to 10 or 19. The check fails for both $s(1919-10)=s(1909)=19$ and $s(1919-19)=s(1900)=10$. Thus $n \leq 6$ and the case $n=6$ is constructed above (1919, 1939, 1961, 1978, 2003, 2008).