Every prime number divide some sum of the first $k$ primes.

Let $S_n=\Sigma^n_{k=1}p_k$, where $p_k$ is the $k$-th prime number.

Conjecture:

$$\forall p\in\mathbb P\exists n\in\mathbb N: p|S_n$$

Verified for the $1000$ first primes. Is there a proof for this result in general?

In the diagram primes are projected on the x-axis and $n$ (as in $S_n$) on the y-axis.

As H.H.Rugh commented there is a stronger conjecture for all positive integers m. Below a table of $n$-records for different $m\in\mathbb Z^+$ (some of them primes):

    m                          n factorization of Sn 
    1 (1)                      1 (2)
    3 (3)                     10 (3,43)
    6 (2,3)                   57 (2,3,5,229)
   12 (2,2,3)                 97 (2,2,3,1879)
   18 (2,3,3)                113 (2,3,3,41,43)
   35 (5,7)                  180 (5,7,2531)
   42 (2,3,7)                305 (2,2,2,3,7,7,239)
   90 (2,3,3,5)              357 (2,2,2,3,3,3,5,367)
  101 (101)                  422 (5,101,1129)
  137 (137)                  861 (2,137,9739)
  163 (163)                  902 (5,7,11,47,163)
  195 (3,5,13)               907 (2,3,3,5,13,2551)
  202 (2,101)               1207 (2,2,19,101,719)
  222 (2,3,37)              1359 (2,2,2,3,3,37,2671)
  252 (2,2,3,3,7)           1683 (2,2,3,3,7,17,37,71)
  326 (2,163)               1765 (2,2,3,23,163,277)
  474 (2,3,79)              2077 (2,2,2,3,5,79,1861)
  504 (2,2,2,3,3,7)         2133 (2,2,2,3,3,3,7,53,233)
  522 (2,3,3,29)            2379 (2,3,3,3,3,3,7,29,239)
  643 (643)                 2529 (2,3,3,11,211,643)
  647 (647)                 3092 (11,647,5791)
  658 (2,7,47)              3353 (2,3,7,43,47,577)
  700 (2,2,5,5,7)           3593 (2,2,5,5,7,103,787)
  817 (19,43)               4683 (2,3,11,19,43,1847)
  995 (5,199)               5329 (2,3,5,17,199,1291)
 1004 (2,2,251)             6415 (2,2,2,251,96643)
 1204 (2,2,7,43)            6533 (2,2,2,2,7,7,31,43,193)
 1459 (1459)                7241 (2,2,3,3,5,5,191,1459)
 1488 (2,2,2,2,3,31)        7307 (2,2,2,2,2,3,31,85909)
 1610 (2,5,7,23)            8079 (2,5,7,23,43,4567)
 1677 (3,13,43)            10171 (2,2,2,3,3,3,13,43,4259)
 1870 (2,5,11,17)          10331 (2,5,11,17,71,4003)
 2035 (5,11,37)            11459 (2,2,3,3,5,11,37,9029)
 2616 (2,2,2,3,109)        11753 (2,2,2,3,3,3,3,3,37,89,109)
 2672 (2,2,2,2,167)        18137 (2,2,2,2,7,167,93047)
 3420 (2,2,3,3,5,19)       21709 (2,2,3,3,3,5,13,19,137,139)
 3830 (2,5,383)            27617 (2,5,53,383,20749)
 4232 (2,2,2,23,23)        38861 (2,2,2,2,3,3,23,23,113189)
 7394 (2,3697)             45381 (2,107,3697,15083)
 7450 (2,5,5,149)          47323 (2,3,3,5,5,7,41,149,677)

$a(n)$, the smallest $k$ such that the $n$-th prime divides the sum of the first $k$ primes, is tabulated at http://oeis.org/A111287. In the comments there, it says:

It follows from a theorem of Daniel Shiu that $k$ always exists. Shiu has proved that if $\gcd(a,b) = 1$ then the arithmetic progression $a, a + b, ..., a + kb, \dots$ contains arbitrarily long sequences of consecutive primes. Since, for any positive integer $b$, there are thus arbitrarily long sequences of consecutive primes congruent to 1 mod $b$, there must be infinitely many $a(n)$ that are divisible by $b$.

To clarify the previous comment: If the sum of the primes up to some point is $s \bmod b$, then we need exactly $b-s$ consecutive primes equal to $1 \bmod b$ to produce a sum divisible by $b$. Hence when there are $b-1$ consecutive primes congruent to $1 \bmod b$, then the sum of primes up to one of those primes will be divisible by $b$. [From T. D. Noe, Dec 02 2009]