China Girls' Mathematical Olympiad

Lemma Let $m$ be the remainder of the sum of the cubes of three positive integers when divided by $9$, then $m\ne 4$ or $5$. Proof: Since any integer can be expressed as $3k$ or $3k\pm 1$ ($k\in \mathbb{Z}$), $$\begin{gathered} (3k{)}^3=9\times 3k^3, \\ (3k\pm 1{)}^3=9\times (3k^3\pm 3k^2+k)\pm 1, \end{gathered}$$ as desired.

If $i=1$, take $n=3(3m-1{)}^3-2$ ($m\in {\mathbb{Z}}^{+}$), then $4$ or $5$ is the remainder of $n$ and $n+28$ when divided by $9$. So, they cannot be expressed as the sum of the cubes of three positive integers. But $$n+2=(3m-1{)}^3+(3m-1{)}^3+(3m-1{)}^3.$$

If $i=2$, take $n=(3m-1{)}^3+222$ ($m\in {\mathbb{Z}}^{+}$), then $5$ is the remainder of $n$ when divided by $9$. So, it cannot be expressed as the sum of the cubes of three positive integers. But $$\begin{gathered} n+2=(3m-1{)}^3+2^3+6^3, \\ n+28=(3m-1{)}^3+5^3+5^3. \end{gathered}$$

If $i=3$, take $n=216m^3$ ($m\in {\mathbb{Z}}^{+}$). It satisfies the conditions: $$\begin{gathered} n=(3m{)}^3+(4m{)}^3+(5m{)}^3, \\ n+2=(6m{)}^3+1^3+1^3, \\ n+28=(6m{)}^3+1^3+3^3. \end{gathered}$$ This completes the proof.