If an integer is not sum of two cubes in integers, then the integer cannot be sum of two cubes modulo every integer. Say some integer $n$ is not sum of two cubes in the integers, then I want to show that there exists $k$ in positive integers such that  $x^{3} + y^{3} \equiv n\pmod{k}$  is not solvable. 
All I seem to know regarding even vicinity of such problems are following:


*

*If $n$ is not sum of two cubes in integers, then exists an integer $\theta(n)$ such that $n\theta(n)$ is sum of two cubes. 

*Any number $n$ is sum of two cubes in integers if and only if following condition is satisfied:
$$\exists m \mid n ,\quad n^{1/3} \leq m \leq 4^{1/3} n^{1/3}$$ such that $( m^{2} - \frac{n}{m} ) = 3l$ and $(m^{2} - 4l)$ is a perfect square. 
May be there is counterexample here, i.e. exists $n$ which is not sum of two cubes yet the congruence equation $x^{3} + y^{3} \equiv n\pmod{k}$ is solvable $\forall k \in\mathbb{N}$. Any help is appreciated. 
 A: $n=20$ is a counterexample (turns out to be the smallest positive one, assuming negative cubes allowed).
Theorem 2.1 in the paper [found by @Mason] states that $x^3+y^3\equiv 20\pmod{k}$ is solvable for each $k$. [UPDATE: Similarly to the answer by Gerry Myerson, we have $20=(1/7)^3+(19/7)^3$, so we're left to deal with $k$ a power of $7$, which is done using Hensel's lemma and the solution $x=6,y=0$ for $k=7$.]
It remains to show that $20$ is not a sum of two integer cubes. Here is an algorithmic recipe. Suppose $n=x^3+y^3=(x+y)(x^2-xy+y^2)$. The second factor is positive, hence $d=x+y$ is a positive divisor of $n$, and we have $n/d=3x^2-3dx+d^2$. This has an integer solution $x$ if and only if $n/d-d^2$ is a multiple of $3$ and the discriminant is a square, i.e. iff $(4n/d-d^2)/3$ is a square of an integer. Examining the divisors of $20$ this way, we're done.
A: $$\left({17\over21}\right)^3+\left({37\over21}\right)^3=6$$ $x^3+y^3=6$ has no solution in integers, positive, negative, or zero (exercise for the reader), but the displayed equation shows there's a solution to $x^3+y^3\equiv6\bmod k$ for every $k$ relatively prime to $21$. 
Now $x^3+y^3\equiv6\bmod3$ has the solution $x=y=0$, and $x^3+y^3\equiv6\bmod7$ has the solution $x=3$, $y=0$. 
This almost takes care of things, but $x^3+y^3\equiv6\bmod9$ has no solution, so this is really close-but-not-quite. 
BUT here's one that works. $$\left({7\over3}\right)^3+\left({11\over3}\right)^3=62$$ $x^3+y^3=62$ has no solution in integers, positive, negative or zero, but the display shows there's a solution to $x^3+y^3\equiv62\bmod k$ for every $k$ relatively prime to $3$. And $2^3+0^3\equiv62\bmod{27}$, together with an application of Hensel's Lemma, takes care of values of $k$ that are powers of $3$. Then the Chinese Remainder Theorem gives solutions for all $k$. 
