Solution:

If $a, b, c$ are the three solutions (not necessarily distinct) of the equation $X^{3}-3 X+n=0$, then
$$X^{3}-3 X+n=(X-a)(X-b)(X-c)=X^{3}-(a+b+c) X^{2}+(a b+a c+b c) X-a b c .$$
It follows that
$$\left\{\begin{array}{l} a+b+c=0 \\ a b+a c+b c=-3 \\ a b c=-n . \end{array}\right.$$
Observing that $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+a c+b c)$, one obtains $a^{2}+b^{2}+c^{2}=6$. If $a, b, c$ are integers, this is possible only if two of the squares are equal to 1 and the third is equal to 4. Therefore we may assume, by symmetry, that $a= \pm 1, b= \pm 1, c= \pm 2$. Since moreover we must have $a+b+c=0$, the only possibilities are $a=b=1, c=-2$ and $a=b=-1, c=2$, from which, respectively, $n=2$ and $n=-2$. By what we have just observed,

$$X^{3}-3 X+2=(X-1)^{2}(X+2), \quad X^{3}-3 X-2=(X+1)^{2}(X-2)$$
and therefore $n=2$ and $n=-2$ are indeed solutions of the problem.