穷举法（Proof by exhaustion）需要考虑所有可能的整数情况。 任何整数 $n$ 被 3 除后的余数只有三种可能：$0$, $1$ 或 $2$。因此，所有整数可以表示为以下三类：

1. $n = 3k$
2. $n = 3k + 1$
3. $n = 3k + 2$ （也可以表示为 $n = 3k - 1$）

题目中学生已经完成了第 1 种情况（$n = 3k$）。要完成该证明，我们只需要考虑剩下的两种情况并分别证明在该情况下 $2n^2 + n + 1$ 不能被 3 整除：

1. 情况二：$n = 3k + 1$。将其代入代数式展开并化简，提取出公因子 3，看余数是否为 0。
2. 情况三：$n = 3k + 2$（或 $n = 3k - 1$）。同理代入展开、提取公因子并分析余数。

最后给出结论：由于所有情况下的表达式都不能被 3 整除，原命题得证。

To complete the proof by exhaustion, we need to consider the remaining two cases for the integer $n$, which are $n = 3k + 1$ and $n = 3k + 2$ (where $k$ is an integer).

Case 2: $n = 3k + 1$

Substitute $n = 3k + 1$ into the expression:

Since $k$ is an integer, $6k^2 + 5k + 1$ is also an integer. Thus, $2n^2 + n + 1$ leaves a remainder of $1$ when divided by $3$, which means it is not divisible by 3.

Case 3: $n = 3k + 2$

Substitute $n = 3k + 2$ into the expression:

Since $k$ is an integer, $6k^2 + 9k + 3$ is also an integer. Thus, $2n^2 + n + 1$ leaves a remainder of $2$ when divided by $3$, which means it is not divisible by 3.

(Alternative for Case 3: $n = 3k - 1$)

which leaves a remainder of $2$ and is not divisible by 3.

Conclusion

Since $2n^2 + n + 1$ is not divisible by $3$ for $n = 3k$, $n = 3k + 1$, and $n = 3k + 2$, which cover all possible integers, it is proved that if $n$ is an integer then $2n^2 + n + 1$ is not divisible by 3.