We claim that all odd numbers are special, and the only special even number is 2. For any even $N>2$, the numbers relatively prime to $N$ must be odd. When we consider $k=3$, we see that $N$ can't be expressed as a sum of 3 odd numbers. Now suppose that $N$ is odd, and we look at the binary decomposition of $N$, so write $N=2^{a_{1}}+2^{a_{2}}+\ldots+2^{a_{j}}$ as a sum of distinct powers of 2. Note that all these numbers only have factors of 2 and are therefore relatively prime to $N$. We see that $j<\log _{2} N+1$. We claim that for any $k \geq j$, we can write $N$ as a sum of $k$ powers of 2. Suppose that we have $N$ written as $N=2^{a_{1}}+2^{a_{2}}+\ldots+2^{a_{k}}$. Suppose we have at least one of these powers of 2 even, say $2^{a_{1}}$. We can then write $N=2^{a_{1}-1}+2^{a_{1}-1}+2^{a_{2}}+\ldots+2^{a_{k}}$, which is $k+1$ powers of 2. The only way this process cannot be carried out is if we write $N$ as a sum of ones, which corresponds to $k=N$. Therefore, this gives us all $k>\log _{2} N$. Now we consider the case $k=2$. Let $2^{a}$ be the largest power of 2 such that $2^{a}<N$. We can write $N=2^{a}+\left(N-2^{a}\right)$. Note that since $2^{a}$ and $N$ are relatively prime, so are $N-2^{a}$ and $N$. Note that $a<\log _{2} N$. Now similar to the previous argument, we can write $2^{a}$ as a sum of $k$ powers of 2 for $1<k<2^{a}$, and since $2^{a}>\frac{N}{2}$, we can achieve all $k$ such that $2 \leq k<\frac{N}{2}+1$. Putting these together, we see that since $\frac{N}{2}+1>\log _{2} N$ for $N \geq 3$, we can achieve all $k$ from 2 through $N$, where $N$ is odd.