How to find remainder? $$a=r\mod (r+1) \ \ \forall r\in\{2,3,4,\dots,9\}$$
Then how do we find $'x'$ if $$a=x\mod 11$$ 
I get $$2a=9\mod11$$ but that does not help.
Please keep solution simple , i don't now number theory.

The above is the crux what I got from the question:

Let $n_1,n_2,... $ be an increasing sequence of natural numbers each of which leaves remainder $r+1$ when divided with $r\in\{2,3,....9\}$. Find the remainder when $n_{2008}$ is divided with $11$.

 A: $$2a\equiv9\pmod {11}\equiv -2\implies a\equiv-1\pmod {11}\equiv10$$

Alternately, as $2\cdot 6=12\equiv1\pmod{11}\implies 2^{-1}\equiv6$
$2a\equiv 9\pmod{11}\implies a\equiv2^{-1}\cdot9\equiv 6\cdot9\equiv 10\pmod {11} $
If $a\equiv r\pmod{r+1}\equiv-1\implies (r+1)$ divides $a+1 \ \ \forall r\in\{2,3,4,\dots,9\} $
If $m_i$ divides $b,$ we know lcm$(m_i)$ will divide $b$
EDIT:  to answer the changed question 
Here lcm$(3,4,\cdots,9,10)=5\cdot 8\cdot7\cdot 9=2520$
So, $n_i\equiv-1\pmod{2520}=2520c_i-1$ for some integer $c_i$s 
For $n_i$ to be natural number $n_i=2520c_i-1>0\implies c>0\implies c_{\text{min}}=1$
So, if we choose, any arbitrary set of increasing positive integers for $c_i,$ we shall get an increasing sequence of natural numbers satisfying the given condition and solution will depend on the choice of $c_i$s
Now if we take $c_i=i$ for $i\ge1,$
$n_{2008}=2520\cdot 2008-1$
$\equiv1\cdot6-1\pmod {11}$ as $2520\equiv1\pmod {11}$ and $2008\equiv6\pmod {11}$
$\implies n_{2008}\equiv5\pmod{11}$
A: Find the least common multiple $L$ of the numbers $3,4,\dots,10$. Then $$a=L-1+kL$$ where $k$ is arbitrary. There is no restriction on $a$ modulo $11$ (unless there's more information on $a$, such as if you take the smallest possible $a$, which is $L-1$). 
A: Use the Chinese reminder Theorem. It looks to me that you have the relations
$$\begin{align}
a&\equiv4\mod{5}\\
a&\equiv6\mod{7}\\
a&\equiv7\mod{8}\\
a&\equiv8\mod{9}\\
\end{align}$$
and all of the other relations give redundant information. There is a unique solution modulo $5\cdot 7\cdot8\cdot9=2520$. Since each of the moduli divide this number, $2519$ is a solution for $a$. So $a\equiv 2519\mod{2520}$. Now what would $a$ be modulo $11$? Anything is possible since $11$ does not divide $2520$. Are you sure this is the question that was asked? If $a$ is supposed to be the smallest positive solution, then $x\equiv2519\equiv0\mod{11}$. If $a$ is the smallest solution in absolute value, then $x\equiv-1\mod{11}$.

Now with the full question posted, do you see how $n_1=2519$? Continue to find $n_{2008}$.
