The answer are all $n$ odd. Let $n=\textstyle\prod_{1\le i\le L}p_i^{e_i}$ be the prime decomposition of an odd $n$. Then, $p_i^{e_i}\mid (a+b)^2$ for all $i$, yielding $v_p(a+b)\ge e_i/2$. So, $v_{p_i}(a+b+c) = v_{p_i}(2a+2b+2c)\ge e_i/2$, yielding $p_i^{e_i}\mid(a+b+c)^2$. Since $i$ is arbitrary, we get the conclusion. Now let $n=2^k \cdot m$ where $m$ is odd and $k\ge 1$. Suppose first $k=2i$ is even. Consider $a,b,c$ such that $a+b = 2^i m$, $b+c=2^i m$ and $c+a = 2^i m$. Then, $a+b+c = 3\cdot 2^{i-1}\cdot m$, so $2^{2i}\nmid (a+b+c)^2$. The case $k=2i+1$ is similar.