The answer is all $n$ that have only prime factors $1 \pmod 4$. It is easy to see that $a+b+c \mid abc+c^2, abc-b^2$, so $a+b+c \mid b^2+c^2$ (in particular, $n$ doesn't have any prime factors $3 \pmod 4$, otherwise we obtain that $\gcd(a, b, c)>1$). Moreover, $a+b+c \mid (b+c)b-c=b^2+bc-c$ and $a+b+c \mid (b+c)c+b=c^2+bc+b$. Subtracting those two, we obtain $a+b+c \mid (c+b)(c-b+1)$. If $\gcd(a+b+c, b+c)>1$, then $p \mid a+b+c$ and $p \mid b+c$ for some prime $p$, which means that $p \mid 2b^2$, so $p=2$, otherwise $p \mid b, c$ and thus $p \mid a$ as well. Hence, $a$ is even and $b, c$ are odd, so $ab+c, ac-b$ are odd but are divisible by $a+b+c$, which is even, contradiction (in particular, a similar argument shows that $n$ is odd). Hence, $a+b+c \mid c-b+1$, so by size, $b=c+1$. Thus, $ab+c=ab+b-1$ and $ac-b=a(b-1)-b=ab-a-b$, so $\gcd(ab+c, ac-b)=\gcd(ab-a-b, a+2b-1)=a+2b-1$, so the condition is equivalent to $n=a+2b-1 \mid ab-a-b \iff a+2b-1 \mid 2b^2-2b+1$. Since $n=a+2b-1$ is odd, the condition rewrites as $n=a+2b-1 \mid 2(2b^2-2b+1)=(2b-1)^2+1$. In the above paragraph, we showed that any working $n$ has prime factors only $1 \pmod 4$. We now show that all such $n$ work. It suffices to find an odd $x$, such that $n \mid x^2+1$ and $x<n$. Let $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$. The congruence $x^2+1 \equiv 0 \pmod {p_i}$ has a solution since $p_i \equiv 1 \pmod 4$, so it has a solution modulo $p_i^{a_i}$ by Hensel's lemma. Thus, by CRT we can find a $x \pmod {2n}$ that is odd and satisfies $n \mid x^2+1$. If $x<n$, we are done, otherwise take $2n-x$ - it is also odd and $2n-x<n$, so it works. This shows that any such $n$ works.

The answer is all $n > 1$ such that all prime divisors $p$ of $n$ have $p\equiv 1\pmod{4}$. First, we show that $n$ must have this property. Obviously, $n \neq 1$, since $n=a+b+c \geq 1+1+1=3$. Furthermore, $n$ is odd, since if we had $n=a+b+c$ was even, then since $\gcd(a,b,c) = 1$, it follows that $(a, b, c)$ is some permutation of $(1, 1, 0)$ modulo $2$. For all such permutations, $ab+c\equiv 1\pmod{2}$, but $n \mid ab+c$, which is absurd since $n$ is even whilst $ab+c$ is odd. So, $n$ must be odd. Furthermore, note that $$n = a + b + c \mid c(ab+c) - b(ac-b) = b^2+c^2,$$so if $p$ is a prime divisor of $n$, then $p\mid b^2+c^2$. Note that neither $b$ or $c$ are divisible by $p$, since that would imply both $b$ and $c$ are divisible by $p$, and since $p\mid n=a+b+c$, it would imply that $a$ is divisible by $p$, so $p\mid \gcd(a,b,c) = 1$, absurd. So, $p\mid b^2+c^2$ and $b, c$ are not $0\pmod{p},$ so $\left(\frac{b}{c}\right)^2 \equiv -1\pmod{p},$ implying that $-1$ is a quadratic residue modulo $p$, so $p\equiv 1\pmod{4}$ or $p=2$. But since $n$ is odd, $p\neq 2$, so we must have $p\equiv 1\pmod{4}$. Thus, if $n$ works, then $n > 1$ and all prime divisors $p$ of $n$ have $p\equiv 1\pmod{4}$. Now, we will show that all such $n$ work. First of all, note that there exists an integer $b$ for which $$n\mid 2b^2-2b+1,$$since for each prime divisor $p$ of $n$, we have that there exists an integer $b$ for which $p\mid 2b^2-2b+1$, since we can take $b \equiv \frac{1+i}{2} \pmod{p}$, where $i\in \mathbb{F}_p$ for which $i^2 \equiv -1\pmod{p}$ ($i\in \mathbb{F}_p$ since $p\equiv 1\pmod{4}$), and then we have that \begin{align*} 2b^2 - 2b + 1 &\equiv 2\left(\frac{1+i}{2}\right)^2 - 2\left(\frac{1+i}{2}\right) + 1 \pmod{p}\\ &\equiv \frac{2\cdot 2i}{4} - (1+i) + 1 \pmod{p}\\ &\equiv 0\pmod{p}. \end{align*}Since $\frac{\mathrm{d}}{\mathrm{db}}(2b^2-2b+1) = 4b-2$, which is only $0\pmod{p}$ when $b\equiv \frac{1}{2}\pmod{p}$, which is not a solution to $2b^2-2b+1\equiv 0\pmod{p}$, as $2\left(\frac{1}{2}\right)^2-2\left(\frac{1}{2}\right)+1 \equiv \frac{1}{2}\pmod{p}$, we have by Hensel's that there exists an integer $b$ for which $p^{\nu_p(n)} \mid 2b^2-2b+1$, so by CRT, there exists an integer $b$ for which $n\mid 2b^2-2b+1$. Note that we may assume $0 < b\leq n$. Note that if $n\mid 2b^2-2b+1$, then $n\mid 2((n+1)-b)^2 - 2((n+1)-b) + 1$, so we actually may assume that $0 < b \leq \frac{n+1}{2}$. Furthermore, note that $b = \frac{n+1}{2} \equiv \frac{1}{2}\pmod{n}$ does not work, since $$2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) + 1 \equiv \frac{1}{2}\not \equiv 0\pmod{p},$$so in fact we must have that $0 < b \leq \frac{n-1}{2}$. Furthermore, $b\neq 1$, since $2\cdot 1^2-2\cdot 1 + 1 \equiv 1\not \equiv 0\pmod{p}$, so $1 < b \leq \frac{n-1}{2}$. Therefore, there exists an $a\geq 1$ for which $n = a+2b-1$. If we let $c = b-1$, then we have that $n = a+b+c$, and that $a,b,c$ are positive integers, since $b\geq 2\implies c\geq 1$, and $a\geq 1$. Furthermore, $\gcd(a,b,c) \mid \gcd(b,c) = \gcd(b, b-1) = 1$, so $\gcd(a,b,c) = 1$. Now, I claim that we have that $$\gcd(ab+c, ac-b) = a+b+c=n.$$To see this, note that $ab+c = ab+b-1$ and $ac-b = a(b-1)-b = ab-a-b$, and $$(ab+c) - (ac-b) = (ab+b-1) - (ab-a-b) = a+2b-1 = a+b+c,$$so it suffices to show that $n\mid ab+c$. To do this, note that $$ab+c \equiv (-b-c)b + c \equiv (-2b+1)b + b-1 \equiv -2b^2+2b-1\pmod{n},$$but since $n\mid 2b^2-2b+1$, it follows that $-2b^2+2b-1 \equiv 0\pmod{n}$. Therefore, it follows that $\gcd(ab+c, ac-b) = n$. So, all conditions are met and so all such $n$ work. Thus, we've shown that all such $n > 1$ such that all prime divisors $p$ of $n$ have $p\equiv 1\pmod{4}$ work and that all other $n$ fail, as claimed. $\blacksquare$