Let $p_1,p_2,\ldots,p_t$ be the prime factors of $n$. If there is an $x$ s.t. $p_i\not |k-x,\ p_i\not |x$, then we can take $a=k-x,b=x$. If $n$ is odd, we can do this because there are at least $3$ residues $\pmod{p_i},\ \forall i$, and we can select $x\not\in\{0,k\}\pmod{p_i}$ (we are sure it exists from the Chinese Remainder Theorem). On the other hand, if $n$ is even, but so is $k$, we select $x\equiv 1\pmod 2$, and for the rest of the primes $p_i$ we do what we did above. So you see, if $n$ is odd, $k$ can be anything we want, which makes me proud of having translated the problem correctly .

Approach by hjbrasch: Induction on the number of prime divisors of $ n$ (not elegant, but it works). As for the induction start assume $ n = p^{t}$. If $ p = 2$ holds, set $ a = 1,b = k - 1$ and remember that $ k$ must be even then. If $ p > 2$ choose any residue $ a$ different from $ 0,k\bmod{p}$ and set $ b = k - a$. As for the induction step write $ n = p^{t}n'$ with $ p$ an odd prime and $ n'$ not divisble by $ p$. Since $ n,n'$ have the same parity, we can assume the existence of integers $ a' + b' = k$, which are both coprime to $ n'$. Consider the quadratic function $ x\rightarrow f(x) = (a' + xn')(b' - xn')$, which is not identical to zero modulo $ p$, since $ n'^{2}$ is not divisible by $ p$. Therefore, there exists $ x$ such that $ f(x)$ is not divisible by $ p$ and we can set $ a = a' + xn', b = b' - xn'$ to complete the proof.