7. Let $f(x)$ be a function defined on the set of natural numbers $\mathrm{N}$, taking values in $\mathrm{N}$, and for $x, y \in \mathrm{N}$, we have $f[f(x)+f(y)]=x+y$. Find $f(1988)$.
7. Let $f(x)=y$, then from the original equation we know $f(2 y)=f[f(x)+f(x)]=2 x$.

Let $f(1)=t$, then we have $f(2 t)=2$.
We will prove $t=1$ by contradiction.
Assume $t=b+1, b \in N$, and let $f(b)=c$, then $2 c=2 b$, so $2 c+2 t=f(2 t)=2$, which leads to $c+t=1$, which is impossible.
Therefore, $f(1)=1$, if $f(k)=k$, then $f(k+1)=f[f(k)+f(1)]=k+1$.
By mathematical induction, for all $x \in N$, we have $f(x)=x$.