This is false if $K>1$ and $n$ is a power of 2.
This is true for $K=1$ for all $n$.
Let’s give the counterexample when $K>1$ and $n$ is a power of 2 first.
When n is a power of 2, you can use the columns of the Hadamard matrix of size $n$ to get $n$ orthogonal vectors from within $C_n$.
Since the vectors $r_i$ form an orthogonal basis and $c$ is a unit vector, we have that:
$1=\sum_i |c \cdot r_i|^2$
Since the average value of $|c_i \cdot r_i|^2$ is $1/n$, at least one component must be at least as much, so there is no vector $c$ all of whom’s components $|c \cdot r_i|$ are less than $1/\sqrt{n}$.
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If we fix $K=1$, then such a $c$ always exists.
Given some $r_i$, let’s show it. Note that if the $r_i$ are linearly dependent, we can just choose an orthogonal vector, so let’s assume they’re independent.
If you think of the $r_i$ as a basis and $R$ as the corresponding matrix, then $C:=R^{-1}$ gives the n $c_i$ vectors for which $r_i \cdot c_j=\delta_{i,j}$. Since the $r_i$ are unit vectors, $|c_i| \geq c_i\cdot r_i / |r_i| =1$.
Let’s loop through the list adding or subtract $c_i$ depending on whether $c_i$ has a positive dot product with the sum so far. Precisely, let $s_0$ be the zero vector. Let $f$ be the sign function (negative to -1, positive to 1, and let use 1 to break ties at 0). Then, for $i\geq 0$, let $s_{i+1}=s_{i}+f(s_i \cdot c_{i+1})c_{i+1}$.
Then by induction we have that $|s_i|^2 \geq i$ since $|s_0|=0$ and $|s_{i+1}|^2=|s_{i}|^2+2f(s_i \cdot c_{i+1})s_i \cdot c_{i+1} + |c_{i+1}|^2 \geq i +0+1$.
Also, as the sum or difference of all the $c_i$ we have that $s\cdot r_i=\sum_{j=1}^{n} f(s_{j-1}\cdot c_j)c_j \cdot r_i=\sum_{j=1}^{n} f(s_{j-1}\cdot c_j)\delta_{i,j}= f(s_{i-1} \cdot c_i)=\pm 1$.
Normalizing $s$ gives a vector with all the desired properties. It would be a unit vector for with $|(s/|s|) \cdot r_i|=1/|s| \leq 1/\sqrt{n}$ for all $i$ as desired.
