12. Let $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$, and for all $m, n \in \mathbf{N}^{*}$, we have $f(f(m)+f(n))=m+n$, then $f(2005)=$
12. 2005 Hint: Let $m=n=1$, then $f(2 f(1))=2$. Suppose $f(x)=y, x, y \in \mathbf{N}^{*}$, then $f(2 y)=2 x$. Let $f(1)=t$, then $t=1$. Otherwise, let $t=b+1, b \in \mathbf{N}^{*}$, and let $f(b)=c, c \in \mathbf{N}^{*}$, then $f(2 c)=2 b$, and $2 c+2 t=$ $f(f(2 c)+f(2 t))=f(2 b+2)=f(2 t)=2, t+c=1$, contradiction! It is easy to prove by induction that $f(n)=n$, in particular, $f(2005)=2005$.