Fix a positive integer $n$. We aim to prove that for any positive integers $a, b, c$ not exceeding $3n^2 + 4n$, there exist integers $x, y, z$ with absolute value not exceeding $2n$ and not all zero, such that $ax + by + cz = 0$.

Without loss of generality, assume $c = \max(a, b, c)$.

Consider the set of integers $x$ and $y$ such that $0 \leq x, -y \leq 2n$ and $x^2 + y^2 > 0$. If any value of $ax + by$ in this set is $0 \mod c$, the norm is at most $2cn$, and we are done. Otherwise, these $4n^2 + 4n$ values are all nonzero modulo $c$. Since $c \leq 3n^2 + 4n$, there must be duplicates modulo $c$.

If $ax_1 + by_1 \equiv ax_2 + by_2 \mod c$, then $a(x_2 - x_1) + b(y_2 - y_1) \equiv 0 \mod c$. If $x_2 - x_1$ and $y_2 - y_1$ are not both positive or both negative, we are done. Otherwise, assume $x_2 > x_1$ and $y_2 > y_1$. We must have $(x_2 + y_2) - (x_1 + y_1) > 2n$ because otherwise $a(x_2 - x_1) + b(y_2 - y_1)$ is positive but at most $c(x_2 - x_1) + c(y_2 - y_1)$, which is at most $2nc$.

The size $x + y$ of $ax + by$ is between $-2n$ and $2n$ inclusive, so there are at most two choices of $(x, y)$ for any fixed residue modulo $c$. If there are three, the largest and smallest have a size difference over $4n$, which is a contradiction.

Since $c \leq 3n^2 + 4n$ and we have $4n^2 + 4n$ values, there must be at least $n^2 + 1$ pairs $((x_1, y_1), (x_2, y_2))$ such that $ax_1 + by_1 \equiv ax_2 + by_2 \mod c$ with $x_2 > x_1$ and $y_2 > y_1$. This gives $a(x_2 - x_1) + b(y_2 - y_1) \equiv 0 \mod c$ with $(y_2 - y_1) + (x_2 - x_1) > 2n$.

If we also have $a(x_3 - x_4) + b(y_3 - y_4) \equiv 0 \mod c$ with $x_3 > x_4$ and $y_3 > y_4$, then:
$$a(x_2 - x_1 - x_3 + x_4) + b(y_2 - y_1 - y_3 + y_4) \equiv 0 \mod c.$$
This implies that the difference has a size of norm at most $2n$ and is divisible by $c$, allowing us to add the appropriate $cz$ and finish.

If all pairs $ax_1 + by_1 \equiv ax_2 + by_2$ have $(x_2 - x_1, y_2 - y_1)$ fixed, then $x_2 - x_1 = A$ and $y_2 - y_1 = B$ with $A + B > 2n$. The range constraints give at most $(2n + 1 - A)(2n + 1 - B) \leq n^2$ choices of $(x_1, y_1)$, which is a contradiction unless $A + B = 2n + 1$.

Thus, $Aa + Bb \equiv 0 \mod c$ with $A + B = 2n + 1$. Since $0 \leq Aa + Bb \leq (2n + 1)c$, it must be a multiple of $c$, and we are done by adding $cz$ unless it is exactly $(2n + 1)c$. This implies $A(c - a) + B(c - b) = 0$, and since $A$ and $B$ are nonzero, either $c = a$ or $c = b$, which are trivial cases.

Therefore, the proof is complete.

The answer is: \boxed{0}.