For $N=4$ we get the projective cubic curve $$ x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3. $$ But this is just the union of $x_1=-x_2$, $x_1=-x_3$, and $x_2=-x_3$, contrary to your requirements. Hence $N \geq 5$. Therefore Gerry Myerson's solution is optimal.

For $N=4$ we get the projective cubic curve $$ x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3. $$ But this is just the union of $x_1=-x_2$, $x_1=-x_3$, and $x_2=-x_3$, contrary to your requirements. Hence $N \geq 5$. Therefore Gerry Myerson's solution is optimal.

Another way to see that $N=5$ can be attained is as follows: for $N=5$, the equations define the smooth, projective cubic surface $X$ given by $$ x_1^3+x_2^3+x_3^3+x_4^3=(x_1+x_2+x_3+x_4)^3, $$ called the Clebsch surface by some authors. Now since $X$ has a rational point, e.g. $(0:0:0:1)$, it has a Zariski dense set of them (for example by Segre--Manin), and therefore it has a rational point $(x_1:x_2:x_3:x_4)$ with all $x_i$ integral and $x_i \neq -x_j$ for $i \neq j$. (Admittedly this argument is phrased in a non-constructive way, but it is in fact easy to write down a birational parametrization of $X$, and so construct a plethora of integral solutions to the system with $N=5$.)