Following Srivatsan's comment, the condition is also sufficient. We can prove it by induction on $n$. The case $n=4$ is easy to deal with.
Suppose that for $n$ the 4-cycle condition $\sigma_{ij}+\sigma_{kl}=\sigma_{jk}+\sigma_{li}$ implies the existence of a solution.
For $n+1$, if the 4-cycle condition is satisfied, by the hypothesis induction there are $a_1,...,a_n$ such that $a_i+a_j=\sigma_{ij},\ i,j=1..n, i \neq j$. Now we are left to pick $a_{n+1}=\sigma_{i(n+1)}-a_i$. For $a_1,...,a_n,a_{n+1}$ to be a solution for the system it is enough to prove that $a_{n+1}$ given by the above relation is the same for every $i=1..n$.
We have $\sigma_{i(n+1)}-a_i=\sigma_{j(n+1)}-a_j \iff \sigma_{i(n+1)}+\sigma_{jk}=\sigma_{j(n+1)}+\sigma_{ik}$, which is true by the 4-cycle condition.