Let be given natural numbers N_1,N_2,N_3,...,N_k such that for every prime p less or equal k set N_1,N_2,N_3,...,N_k does not contain all reminders modulo p. Is it right that there exists number X such that all X+N_1,X+N_2,X+N_3,...,X+N_k are prime? I think it must follow from some theorems about prime numbers in arithmetical progression.

8$\begingroup$ If you could solve the case N_1=1, N_2=3 I'd be very interested ... $\endgroup$ – gowers Mar 9 '10 at 11:21

$\begingroup$ I don't think the OP is asking for infinitely many. I can settle your case by exhibiting X=2, can't I? $\endgroup$ – TonyK Mar 9 '10 at 11:25

$\begingroup$ Even this variant is not going to work easily: simply look for pairs of consecutive primes arbitrarily far apart from each other to find different twin primes. $\endgroup$ – damiano Mar 9 '10 at 11:29

8$\begingroup$ Yesperhaps one should instead do $N_1=1$, $N_2=3$, $N_3=N+1$ and $N_4=N+3$ for some superlarge $N$ congruent to 0 mod 6. This definitely forces infinitely many twin primes. $\endgroup$ – Kevin Buzzard Mar 9 '10 at 11:36
(This question has been killed in the comments, but it is still lacking the useful pointers.)
This is a weak form of the HardyLittlewood Conjecture which moreover predicts an asymptotic density for the number of such prime ktuplets. Special cases of this include: twin primes, cousin primes, sexy primes, prime quadruplets, quintuplets, and sextuplets. While your conjecture is much weaker than HardyLittlewood, Kevin Buzzard's trick in the comments shows that it globally implies the infinitude of prime ktuplets for any admissible pattern.
As far as I know, the infinitude of prime ktuplets is an open problem for all fixed admissible patterns with k ≥ 2. Note that the GreenTao Theorem falls short of proving any instance of this since the step size of the arithmetic progressions is not fixed. (Even the TaoZiegler Theorem falls short since the polynomials are required to have vanishing constant term.)

3$\begingroup$ It has been conjectured that, for any $k>0$ and $n>2$, there are at least as many primes between 2 and $n$ (inclusive) as there are between $k+2$ and $k+n$. About 35 years ago, Hensley and Richards proved that this conjecture is incompatible with the prime $k$tuples conjecture. It's the only example that comes to my mind of two plausible conjectures being proved incompatible. It's generally believed now that prime $k$tuples is true and the other conjecture is false. $\endgroup$ – Gerry Myerson Mar 9 '10 at 23:24