There exist infinitely many numbers which... 
There exist infinitely many numbers which are sums of seventy primes and divide exactly an odd number which itself is a sum of seventy primes.

How does one go about proving this?
 A: Since any odd number greater than $5$ is the sum of three primes, any number big enough is the sum of seventy primes, and there is very little to prove.
A: We wish to show that there are infinitely many natural numbers $n$ such that $n$ is the sum of $70$ primes, and such that $n$ divides an odd number which is also a sum of $70$ primes.
As noted by others, we can just take $n$ to be the sum of $70$ primes one of which is $2$ and the remainder of which are odd, and then because trivially $n \mid n$, we have that $n$ satisfies the conditions of the problem.
This isn't very interesting however. It turns out that the result is still true if we specifically require that $n \mid m$ for some odd number $m \neq n$ such that $m$ is the sum of $70$ primes, and even remains true if we require that $n$ divides infinitely many such $m$.
Indeed, let $M$ be the sum of your favourite $69$ primes, one of which is $2$. (So that $M$ is an even number) We will look for numbers $n$ such that $n = M + p$ and $n \mid M + q$ for some primes $p$ and $q$. (Where $p \neq q$ so that we exclude the case $n = m$)
But this becomes easy if we use Dirichlet's Theorem on primes in arithmetic progressions. (Which may be slightly overpowered) 
Indeed, consider any prime $p$ such that $p$ does not divide $M$. Then by Dirichlet's Theorem, there are infinitely many primes $q$ such that $q = k(M + p) - M$ for some integer $k$. But for any such prime $q$, we clearly have that $M + p \mid M + q$, and so we are done.
A: Perhaps you want the "divide" to be proper, i.e. exclude cases where the first number and the number it divides are the same.
In fact, let $n$ be any odd positive integer (it can be the sum of $70$ primes if one of those primes is $2$ and the others are odd).  I claim that there are infinitely many  odd sums of $70$ primes that are divisible by $n$.
Namely, let $m = \sum_{j=1}^{70} p_j$ where $p_1 = 2$, the other $p_j$ odd primes, and for $j = 2$ to $35$, $p_j \equiv 2 \mod n$ while for $j = 36$ to $70$, $p_j \equiv -2 \mod n$.  Dirichlet's theorem on primes in arithmetic progressions tells us that there are infinitely many possible choices for the $p_j$.
A: Choose 69 odd primes and 2. The sum will be odd, $69\cdot 1 +0\equiv 1 \pmod{2}$, so we have a number which is the sum of seventy primes and divides exactly an odd number which itself is a sum of seventy primes (namely it divides itself exactly). Since we can pick an arbitrarily large odd prime, we can find such numbers that are arbitrarily large. Thus there must be infinitely many of them.
A: There are infinitely many primes. Let $p_i$ denote the $i$'th prime (w.r.t. the usual ordering on $\mathbb N$). Let $P = \{p_1, \cdots, p_{69}\}$ denote the first 69 primes. $p_1 = 2$ and other 68 are odd, thus $\sum P$ is even. Now for any $i \ge 70$ consider $x_i = p_i + \sum P$. Since the sum of an odd number of odd numbers is again odd (and each $p_i$ is odd), we have that $x_i$ is odd. Obviously, we have $x_i|x_i$, and $x_i$ is an odd number which is the sum of seventy primes, so $x_i$ satisfies the condition you require. The function $\mathbb N_{\ge 70} \to \mathbb N$ which has $i \mapsto x_i$ is injective, thus there are infinitely many such numbers $x_i$.
