1. i=1, $n=3m^{3}-1$ $m^{3}+m^{3}+m^{3}\in (n,n+2)$. 2. i=2 $n=3m^{3}+9m^{2}+27m+14$, then $(m+2)^{3}+(m+2)^{3}+(m-1)^{3}=n+1,(m+3)^{3}+m^{3}+m^{3}=n+15.$ 3. i=3, I find $n=3m^{3}+27m^{2}+195m+332$, then \[N1=(m+6)^{3}+(m+5)^{3}+(m-2)^{3}=n+1, N2=(m+7)^{3}+(m+2)^{3}+m^{3}=n+19, N3=(m+8)^{3}+(m+1)^{3}+m^{3}=n+181.\]

Our "Maths discussion group" seems solve the case of $i=1,3$: First,a lemma,$a^{3}+b^{3}+c^{3}$ can't be $4,5(mod 9)$ Brief proof:$x^{3}=0,1,-1 (mod9)$ and try all case of $a,b,c$ in mod 9 and result is out. For $i=1$,take $n=(9a+1)^{3}+(9b+1)^{3}+(9c+1)^{3}=3 (mod9)$ then $n+2=5(mod9)$ and $n+28=4(mod9)$,both can't be sum of 3 cubes. For $i=3$, take $n=k^{3}+(6k)^{3}+(8k)^{3}$, then $n+2=(9k)^{3}+1+1$ and $n+28=(9k)^{3}+27+1$ Also, one of our "group member" think that if we can prove there are infinite $d$ such that $d^{3}=a^{3}+b^{3}+c^{3}$ have no integral solution, then the case of $i=2$ is also solved

ychjae, i=2 is solved. Observed that 5^3+5^3 = 6^3 + 2^3 + 26 Therefore, we can take n+28 = (3k-1)^3 +5^3+5^3 n+2 = (3k-1)^3 + 6^3 + 2^3 n =(3k-1)^3 + 6^3 + 2^3 -2=(3k-1)^3 + 222 which can't be the sum of three cubes by the lemma. By the way, i think Rust's solution is wrong.

For $i = 1$ take $n = 1^3+1^3+(3k+1)^3$ for any $k \in \mathbb{N}.$ Then, using the fact that any sum of three cubes is congruent to $\pm 3, \pm 2, \pm 1, 0$ (mod $9$), we know that $n+2, n+28$ are not sums of three cubes. For $i = 2$ take $n = (3k+1)^3 + 10^3 + 18^3$ for any $k \in \mathbb{N}.$ In this way, $n+28 = (3k+1)^3 + 1^3+19^3$ and $n+2 \equiv 4$ (mod $9$), so that $n, n+28$ are sums of three cubes whereas $n+2$ is not from the earlier fact. For $i = 3$ take $n = 216k^3.$ In this way, $n = (3k)^3 + (4k)^3 + (5k)^3, n+2 = (6k)^3+1^3+1^3, n+28 = (6k)^3+1^3+3^3.$ $\square$