7. $n$ is a positive integer, $f(n)=\sin \frac{n \pi}{2}$. Then
$$
f(1991)+f(1992)+\cdots+f(2003)=
$$
$\qquad$
7. -1 .

It is easy to know that $f(1991)=-1, f(1992)=0, f(1993)=1$, $f(1994)=0, \cdots, f(2003)=-1$. Therefore, $f(1991)+f(1992)+\cdots+f(2003)=-1$.