Assume we have a 21 consecutive Niven numbers, with the smallest one being $ a$. Let $ s(x)$ denote the sum of digits of $ x$. Now if we have two odd Niven numbers $ b$ and $ b+10$ whose tens digits differ by one (as opposed to them being 9 and 0), then $ s(b+10)=s(b)+1$. Therefore one of $ s(b),s(b+10)$ is even, which is a contradiction since $ b$ and $ b+10$ are odd. Such a situation clearly arises if no number in our list in divisible by $ 100$; we could take the pair $ (a,a+10)$ or $ (a+1,a+11)$. Therefore some number in our list is divisible by $ 100$; call it $ c$. The list cannot have both $ c+1$ and $ c+11$, so $ c+11$ is not in the list. Similarly, we cannot have both $ c-1$ and $ c-11$, so $ c-11$ is not in the list. This restricts our list to be $ c-10$ through $ c+10$. Now for $ 0\le i\le 9$, we have $ s(c+i)=s(c)+i$. Then one of $ s(c),s(c+1),...,s(c+9)$ is divisible by 10. This number must be $ s(c)$ since none of $ c+1,...,c+9$ are divisible by $ 10$. Since $ s(c)$ cannot be $ 0$ (as $ 0$ dividing a number is not well-defined), we must have that $ s(c)\ge 10$. Also note that $ s(c+1)=s(c+10)$. Then $ s(c+1)|c+1$ and $ s(c+1)|c+10$, so $ s(c+1)|(c+10)-(c+1)=9$ which means that $ s(c+1)\le 9$. However, this means that $ 9\ge s(c+1)=s(c)+1\ge 10+1=11$ which is a contradiction.