The answer is NO, it is not possible. Let denote the progressions with $P_i,\, i=1,2,\cdots,2008 , P_i = \left\{a_i + kd_i \right\}_{k=0}^{\infty}$. $P_i \cap P_j = \emptyset \, <=> \, \, \not\exists n,m \in \mathbb{N}, \, a_i - a_j = n d_j - m d_i \, <=>\, \gcd(d_i, d_j)$ don't divide $a_i - a_j\,\, \Rightarrow $ (1) $ \gcd(d_i, d_j) > 1 , \, \forall i,j ,\, i \neq j $. Let consider $ \{kd_1\}_{k=1}^{\infty}$. From the problem's clause, it follows, that from some point on, every such number belongs to some progresion. Let $ k d_1 \in P_j$ for some $k,\,j \,\, \Rightarrow $ (2) $ k d_1 = a_j + n d_j$. Let $ d=\gcd(d_1, d_j)$ From (1) it follows $ d>1 $. Using (2) we get $ d / a_j $. But $ d/a_j $ and $ d / d_j $ contradicts the fact that $P_j$ contains a prime number.

dgrozev wrote: Let $ d=\gcd(d_1, d_j)$ From (1) it follows $ d>1 $. Using (2) we get $ d / a_j $. But $ d/a_j $ and $ d / d_j $ contradicts the fact that $P_j$ contains a prime number. Your idea is very correct, except a small problem, $ d/a_j $ and $ d / d_j $ but $P_j$ can still have a prime number, for instance $a_j$ is a prime number and $d$ is, of course, a prime too. Here's my solution I hope that it's correct (my idea is almost the same as yours, except a small lemma) Suppose the contrary, which means we can find such 2008 sequences, let their differences are $d_1,d_2,...,d_{2008}$ and their first terms are $a_1,...,a_{2008}$, respectively Lemma: There exists at least one number of $d_1,d_2,...,d_{2008}$, which is less than or equal to 2008 Proof: Assume the contrary, or WLOG, $2008<d_1\leq d_2\leq ... \leq d_{2008}$ From the condition, there must exist a constant natural number M such that for all $n\geq M$ then $n$ belongs to at least one of 2008 sequences given above (*). Choose a number k large enough, such that $a_1+kd_1\geq M$ then we consider the interval $(a_1+kd_1,a_1+(k+1)d_1)$ there are exactly $d_1-1$ natural numbers in that interval, and by the minimum of $d_1$ once can easily check that there exists at most one number from each sequence (except from sequence $(a_1,d_1)$ of course) between the interval (or there must exists $d_j<d_i$ which is a contradiction), Now, we conclude that the number of natural numbers (which are not in any sequence) in $(a_1+kd_1,a_1+(k+1)d_1)$ is at least $d_1-1-2007=d_1-2008>0$ which is a contradition (see (*)), hence the lemma is proven Now everything is almost the same as your solution, WLOG, suppose that $d_1\leq 2008$ we consider $a_1+kd_1$ if $(a_1,d_1)=d>1$ then we from the condition of existence of prime number we conclude that $d \in P$ and $a=p$ thus $p>2008$ which implies $d_1>2008$ (since $p|d_1$) a contradiction So $(a_1,d_1)=1$ we consider a number $L>M$ (M has been presented in the lemma) and $d_1|L$ then by the definition of $M$ we conclude that $L$ must be in one sequence, named $(a_i,d_i)$ Then $d_1|a_i+d_i.k$ for some $k$, if $(d_i,d_1)=1$ then the equation $a_i-a_1=ud_i-vd_1$ has solution $u,v >0 $ and $u,v$ are natural number, (a well-known problem) then $(a_i) \cap (a_1)$ is not an empty set, which is again a contradiction! Thus $(d_i,d_1)=d>1$ then $d|a_i$, similarly, we conclude $d \in P$ and $a_i=p$ then $p>2008$ or $d_1>2008$ (since $p$ is a divisor of $d_1$) which makes the assumption is wrong then the answer is NO

Here is another solution. We claim that the answer is $\boxed{no}$. Assume, for contradiction, that it was possible. Consider the common differences $d_1, d_2, \cdots, d_{2008}$ of the arithmetic progressions. It's easy to see (density) that $$\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_{2008}} = 1. \qquad (1)$$From $(1)$, we know that at least one of the $d_i$'s is even, as otherwise multiplying by $d_1d_2 \cdots d_{2008}$ would give a contradiction. Notice that any of the arithmetic progressions with even common difference must take on only odd values, since else this progression would contain no odd prime. Therefore, the subset of the evens which are in some arithmetic progression has density $$\sum_{i=1}^{2008} f(i),$$where $f(i) = \frac{1}{d_i}$ if $d_i$ is odd and $0$ else. Since there is some $d_i$ which is even, this quantity is clearly less than $1$, implying that there are infinitely many evens in no arithmetic progression. This contradicts the assumption of the problem. $\square$