Write the primes in increasing order

  P1 = 2 , P2 = 3 , P3 = 5 , … , Pn , Pn+1 , …

and put

  Sn = Pn + Pn+1 .

The problem asks for the least natural number k such that only finitely many Sn are divisible by k.

1. What we can prove at present
   • Every Sn with n ≥ 2 is even, so k = 1 or 2 certainly will not work.
   • Using the elementary argument you gave (it can be rewritten a little more cleanly in terms of residues) one sees that 3, 4 and 6 also fail: there are infinitely many n with 3|Sn, and therefore with 4|Sn and with 6|Sn as well.

These are essentially the only results that can be proved with present–day techniques. As soon as k contains a prime factor ≥ 5 nothing is known.

2. Why nothing more is known
   To force Sn ≡ 0 (mod k) when k contains a prime factor p ≥ 5 one has to produce infinitely many consecutive primes having one of the pairs of residues (r , −r) modulo p with r ≠ 0. For instance for p = 5 the required pairs are (1,4) or (2,3) (in either order). Nothing at all is known about the existence of infinitely many consecutive primes of any prescribed pair of residues. In particular, even the following famous special case is still open:

  Twin-prime problem Are there infinitely many consecutive primes with the   difference 2?

The twin–prime conjecture is the case “difference = 2”. Your problem for k = 5 (which asks for infinitely many consecutive primes whose residues are (1,4) or (2,3) modulo 5) is of the same type, and is at least as hard as the twin–prime conjecture – indeed, either kind of pair of residues would give a difference that is ≡ ±3 (mod 5), and to show that one of those differences occurs infinitely often is already completely out of reach at present.

Modern progress on bounded gaps between primes (Zhang, Maynard, Tao, 2013-…) guarantees that there are some integers g ≤ 246 such that Pn+1 − Pn = g happens infinitely often, but the proofs do not identify a single concrete g, let alone allow us to prescribe at the same time the residue classes of the two primes modulo an independent number k.

3. What would happen if the Hardy–Littlewood prime-tuple conjecture were true
   The Hardy–Littlewood conjecture predicts that every admissible pattern of prime linear forms occurs infinitely often. The pair of linear forms

  n ↦ kn + r ,  n ↦ kn − r   (0 < r < k, gcd(r , k) = 1)

is admissible, and the conjecture therefore implies that for each k there are infinitely many n for which the two displayed numbers are consecutive primes; but their sum is 2kn and is clearly divisible by k. Hence under the conjecture no such least k exists at all: for every k there would be infinitely many Sn that are divisible by k.

4. Present state of the problem
   Unconditionally we can rule out k = 1,2,3,4,6 (your argument).
   For every other k we do not know whether there are finitely or infinitely many Sn divisible by k. In particular we cannot even decide the next candidate k = 5. Therefore

  With current knowledge the “least k” cannot be determined; the question is   open, and may well be as hard as the twin-prime conjecture.

So the best we can say today is

  k ≥ 5, and no concrete value of k ≥ 5 can presently be shown to have the   required finiteness property.