Not as hard as it looks

We require $1 \le n \le 1000$ such that $n^{3} - 4n = n(n^{2} - 4) = n(n - 2)(n + 2)$ be divisible by $144 = 2^{4}3^{2}.$

Now if $n$ is odd then none of $n, (n - 2), (n + 2)$ are divisible by $2,$ and hence their product cannot be divisible by $2^{4}.$ If $n$ is even, then all of these factors are divisible by $2$ and either one or two are divisible by $4.$ In either case their product will then be divisible by $2^{4}.$

Precisely one of $n, (n - 2), (n + 2)$ will be divisible by $3,$ so for the product of the three terms to be divisible by $3^{2}$ we will need one of them to be divisible by $9.$

So combining these observations, we see that for each $n$ that is an even multiple of $9,$ i.e., $n = 18k,$ each of $n = 18k - 2, 18k$ and $18k + 2$ will be such that $n(n - 2)(n + 2)$ will be divisible by $144.$ For $k = 0$ we have just one of the values in the desired range, (namely $n = 2$ ), but since $18*55 + 2 = 992$ we see that for $1 \le k \le 55$ each of these $3$ expressions will yield valid solutions.

There will thus be a total of $1 + 3*55 = \boxed{166}$ solutions.