Adding coprimes to get a coprime These questions arose in my research. Let $n$ be odd and let  $\mathcal{C}_n$ be the set of integers less than $n$ which are coprime with $n$. 
Question 1: For each integer $\ell: 0 \leq \ell < n$ can we always find a subset $\mathcal{S} \subset \mathcal{C}_n$ such that
$$ \ell+2\cdot\sum_{c \in \mathcal{S}}c \quad (\textrm{mod } n)$$
is coprime with n?
That is, can we always find some set of numbers coprime with $n$ such that twice their sum added to $\ell$ is coprime with $n$? Can we always get $|\mathcal{S}| = 1$ when $\ell$ is not coprime with $n$?
Question 2: What is the smallest set $\mathcal{B_n} \subset \mathcal{C}_n$ such that Question 1 holds with $\mathcal{C}_n$ replaced by $\mathcal{B}_n$?
 A: This is extended Douglas Zare's comment explaining why we may take $|S|=1$. We prove that for any residue $\ell$ modulo $n$ there exist $a,b$ coprime to $n$ such that $\ell+a \equiv b \pmod n$. Then we may take $S=\{a/2\}$. 
For any prime $p$ which divides $n$ choose remainder $a_p$ modulo $p$ such that neither $a_p$, nor $a_p+\ell$ is divisible by $p$. This is clearly possible, there are (at least) $p-2$ ways to do it. Then by Chinese Remainders Theorem there exists $a$ congruent to $a_p$ modulo each $p|n$. Use it.
As for the question 2, we may take ${\mathcal B}_n$ with $k$ elements, where $k$ is the number of distinct prime divisors of $n$ (this is in general better than powers of 2 not exceeding $n$). For any $p|n$ choose a number $b(p)$ such that $2b(p)$ gives remainder 2 modulo $p$ and 1 modulo other prime divisors of $n$. Fix $\ell$ and for $m=0,1,\dots$ denote by $f(m)$ the number of primes $p|n$ for which $p|\ell+m$. 
Lemma. Assume that $f(m)\leqslant m$ and $f(m+1)\leqslant k-m$. Then there exists a set $A$, $|A|=m$, of prime divisors of $n$, such that $\ell+\sum_{p\in A} 2b(p)$ is coprime to $n$. 
Proof. Denote $x=\ell+\sum_{p\in A} 2b(p)$. Then $x\equiv \ell+m+\chi_A(p) \pmod p$ for all prime $p|n$. So, there are $f(m)$ primes which have to be included to $A$ and $f(m+1)$ other primes which can not be included. Clearly such $A$ exists exactly when $f(m)\leqslant m$ and $f(m+1)\leqslant k-m$.
Now we need to find appropriate $m$. It is not a big deal, since we always have $f(m)+f(m+1)\leqslant k$ and $f(m)+f(m+2)\leqslant k$. 
If $k$ is even, $k=2q$, we examine $m=q-1$ and $m=q$. Assume that both do not work. If $f(q)\leqslant q$, then $f(q-1)\geqslant q+1$, $f(q+1)\geqslant q+1$, a contradiction. If $f(q)\geqslant q+1$, then both $f(q+1)\leqslant q-1$ and $f(q+2)\leqslant q-1$ and we may take $m=q-1$.
If $k=2q-1$, consider $f(q)$. If $f(q)\leqslant q$, then $f(q-1)\geqslant q$ (else $m=q-1$ works) and $f(q+1)\geqslant q$ (else $m=q$ works). A contradiction. If $f(q)\geqslant q+1$, then $f(q+1)\leqslant q-2$, $f(q+2)\leqslant q-2$, so $m=q+1$ works.
A: To make up for my  not-well-thought-out comment to Fedor Petrov's answer, here is an answer to show $B$ can have $O((\log k)^2)$ elements, where $k$ is the number of distinct prime factors of $n$.  If one pursues the literature, one can take this down to $O(\log k)$ elements.
$n$ being odd, $2$ and its powers are coprime to $n$, with $d=(n+1)/2$ being a multiplicative inverse mod $n$.  Now $\gcd(l+2b,n)=\gcd(dl+2db,n)=\gcd( c+b,n)$, so we just need to find the greatest distance between an arbitrary number $c$ and the next largest coprime integer $c+b$.  More simply put, $l+2i$ is coprime to $n$ for at least one $i$ greater than $0$ and less than a value which I will call $g(n)+1$.
This is related to Jacobsthal's function $g(n)$, of which I know a little something.  One has $g(n) \leq k^{A + C\log\log k}$, where $A,C \lt 4$, which can be found in an arxiv preprint (cf MathOverflow 37679), which lets us take $B$ to be the powers of 2 from 1 up to $\log(g(n))$, giving fewer than $O((\log k)^2)$ elements.
Gerhard "Sorry For Not Thinking Earlier" Paseman, 2016.01.13
