jmc

The given polynomial has degree $4,$ so either it is the product of a linear term and a cubic term, or it is the product of two quadratic terms. Also, we may assume that both terms have leading coefficient $1.$

In the first case, the linear term must be of the form $x-a,$ so the polynomial must have an integer root $a.$ That is, $a^4-na+63=0$ for some integer $a.$ Since $n>0,$ this is impossible when $a\le 0,$ so we must have $a>0.$ Then $$n=\frac{a^4+63}{a}=a^3+\frac{63}{a}.$$Testing various positive divisors of $63,$ we see that $n$ is minimized for $a=3,$ giving $n=3^3+\frac{63}{3}=27+21=48.$

In the second case, let $$x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$$for some integers $a,b,c,d.$ Comparing the $x^3$ coefficients on both sides shows that $a+c=0,$ so $c=-a.$ Then, comparing the $x^2$ coefficients, we get $$b+ac+d=0⟹b+d=a^2.$$We also have $bd=63,$ looking at the constant terms. The only possibilities for $(b,d)$ are $(b,d)=(1,63),(7,9).$ Then the corresponding values of $a$ are $a=\pm 8,\pm 4,$ giving $n=\pm 496,\pm 8.$ Therefore the least value for $n$ is $\boxed{8}.$