11. Given the function defined on the set of non-zero natural numbers
$$
f(n)=\left\{\begin{array}{ll}
n+2, & n \leqslant 2005 ; \\
f(f(n-4)), & n>2005 .
\end{array}\right.
$$

then when $n \leqslant 2005$, $n-f(n)=$ $\qquad$
when $2005<n \leqslant 2007$, $n-f(n)=$
$\qquad$
11. $-2,0$.
When $n \leqslant 2005$, we have
$$
n-f(n)=n-(n+2)=-2 \text {. }
$$

When $2005<n \leqslant 2007$, we have
$$
2001<n-4 \leqslant 2003,2003<n-2 \leqslant 2005 \text {. }
$$

Thus, $n-f(n)=n-f(f(n-4))$
$$
=n-f(n-2)=n-n=0 \text {. }
$$