Covering $\mathbb{N}$ with prime arithmetic progressions For every prime $p_i>2$ choose a $k_i\ge p_i$ , $k_i \in \mathbb{N}$ and take the arithmetic progression $A_i=k_i+np_i$ $n \ge 0$ . Is there any choice of the $k_i's$ such that $|\mathbb{N} \backslash \bigcup A_i | < \infty $ ? 
ADDED Does it makes any diferrence if we omit some other prime number (not 2)?
 A: Here is a graph generated by the first 7500 steps of the method described above. At each stage it finds the smallest uncovered integer $m$ greater than 100 and covers it with a progression $r_i+np_i$ for $n\ge1.$ The last few primes chosen and corresponding  $m$ covered are 
$ [74099, 94245], [74297, 94263], [75329, 94281], [77893, 94283] [74903, 94296],$ $  [77479, 94334], [77611, 94355], [77659, 94361], [74897, 94371], [77977, 94403]$
At this stage the gap $m-p_i$ appears to be around $16500$ for $m \bmod{3}=1$ and $19500$ for $m \bmod{3}=2$
The graph itself shows the number of unused odd primes $p \lt m$ at each stage. Starting after step 1000 or so it seems to increase pretty reliably at an average rate of slightly over $0.23$ for each step. 

A: If you omit the condition that $k_i\ge p_i$, then here is an answer: for every integer $n$ there is some odd prime dividing $2n+1$. So choosing the $k_i$'s so that $2k_i+1\equiv0\pmod{p_i}$ provides a complete covering of the integers (by the congruence classes $\frac{p-1}2\pmod p,\;p>2$). Note that with the condition that $k_i\ge p_i$ this does not cover the numbers $(p-1)/2$.
A: I like Aaron Meyerowitz's efforts and think his and similar methods deserve further study.  I want to post my skepticism as a counter, and hope that something will arise from the contrast.  I do not consider this post as being an acceptable answer though.
The problem is essentially a shifted sieving problem.  After taking the first $n$-many (finitely) primes $q_i$ with offsets $r_i$, one has an eventually periodic pattern of uncovered integers which repeats with period $Q_n = \prod_{i \leq n} q_i$, which contains $U_n = \prod_{i \leq n} (q_i - 1)$ uncovered numbers in each period, and has the first period starting somewhere near $M_n = \max_{i \leq n} r_i$.  
If the $q_i$ are the primes in ascending order, we have (Mertens) that $U_n$ is 
$O(Q_n/\log(q_n))$, which is (roughly) about $n$ times as many primes in the interval
$(M, M + Q_n)$ when $n$ gets large, especially when $n$ is comparable to the largest
integer $M$ allowed to be uncovered.
If the distribution of coprimes to $Q_n$ were amenable to being nicely covered by arithmetic progressions of primes less than $q_n$, I might share Aaron's confidence.
However, each later prime $q$ used is itself coprime to $Q_n$, and with small deviation will cover only about $1/q$ of what needs to be covered.  I suspect that when $n$ gets
to be about $Q_{24}/2$ using Aaron's sequence $Q_i$, he will run short on primes.  It
might be prudent to try more extensive simulations which leave no numbers greater than
50 uncovered.
Gerhard "Saying As I Feel It" Paseman, 2011.11.18 
