We will prove the result for any arbitrary natural $m$ without using $2008$ .First observe that for any odd prime $p$ there exists some $r$ such that $1 \le r \le p$ and the congruence $k^2+k \equiv r (mod p)$is unsolvable .If otherwise for all such $r$ this should be solvable $\Rightarrow k^2+k\equiv r(modp) \Rightarrow$ for all $r$ with $1\le r \le p$ there exists $k$ such that $(2k+1)^2\equiv 4r+1( modp)$ Observe that the set of numbers of the form $4r+1$ forms a complete set of residues $(mod p)$ Which forces us to conclude that for all $r$ the congruence $l^2\equiv r(modp)$ is solvable .But for any prime $p$ there are exactly $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ non residues .. Hence a contradiction.If $p=2$ we have $k^2+k$ is always even.Hence for any prime $p$ there exists $r$ such that $k^2+k\equiv r(modp)$ is unsolvable.Let $p_1 ,p_2,....,p_i$ be the set of primes less than $m$ and let $b_1,b_2,....,b_i$ be the numbers such that the congruence $k^2+k\equiv b_j(modp_j)$ is unsolvable.By chinese remainder theorem there exists $n$ such that the congruences $n \equiv -b_1(modp_1),......,.n\equiv -b_i(modp_i)$ has a simultanequs solution It is easy to observe that $k^2+k+n \equiv k^2+k-b_j(modp_j)$.Since for any $j,, p_j\nmid k^2+k-b_j$ this $n$ satisfies the given conditions