Don't use modulo here!

Let $\large{\frac{n^3-3}{n-3} = k \in \mathbb{Z}}$ . Then: ${\Rightarrow \frac{n^3 -9n^2 + 27n -27 +9n^2 - 27n + 27 -3}{n-3} = k \in \mathbb{Z} \\ \Rightarrow \frac{(n-3)^3 +9n(n-3) +24}{n-3} = k \in \mathbb{Z} \\ \Rightarrow \frac{24}{n-3} = k - (n-3)^2 -9n\in \mathbb{Z} }$ Which means that $n-3|24$ , thus $n-3$ is a divisor of $24$ . Hence: ${(n-3) \in \big\{\pm1, \pm2, \pm3, \pm 4, \pm6, \pm8, \pm12, \pm24\big\} \\ \Rightarrow \boxed{n \in \Big\{-21, -9, -5, -3, -1, 0, 1, 2, 4, 5, 6, 7, 9, 11, 15, 27\Big\}}}$

Let $a = n-3$ . Now we just need values of integer $a$ such that $a | (a+3)^3 - 3$ . Since $a$ divides all the $a$ terms after expansion of the polynomial, we only need $a$ such that $a|24$ . There are 16 values of $a$ that satisfy this and none of them are $a=0$ so the answer is $16$ .

A simpler way to solve this:

We know that $a-b\mid a^3-b^3$ since $a^3 - b^3 = (a-b)(a^2+ab+b^2)$

So we get $n-3 \mid n^3 - 27$ for all integers $n \ne 3$ . We can rewrite $n^3-3$ as $(n^3 - 27) + 24$ . We know that $n-3$ divides the first term and for $n -3$ to divide the whole expression we must have: $n-3\mid 24$

24 has 8 factors and since we are looking for integers we include the negative factors, we have the answer as 16.

Given $n-3 | n^3 - 3$ but the polynomial division leaves a remainder of $\frac{24}{n-3}$ with quotient $n^2+3n+9$ .

Note This is easy to see as $n^2 - 27$ is the closest polynomial from $n^2-3$ that is divisible by n-3

So this problem reduces to finding all $n \in \mathbb{Z}$ such that $n - 3\mid 24 \Rightarrow n - 3 \mid 2^3 \bullet 3^1$

So no of integers $= 2 \bullet (3+1) \bullet (1+1)$