10 Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$ and for any $x \in$ $\mathbf{R}$, we have
$$
\begin{array}{l}
f(x+5) \geqslant f(x)+5 \\
f(x+1) \leqslant f(x)+1 .
\end{array}
$$

If $g(x)=f(x)+1-x$, then $g(2002)=$
Analyze: First determine $f(2002)$ from the given conditions, then we can find $g(2002)$.
Solution: From the given conditions, we have
$$
\begin{aligned}
f(x)+5 & \leqslant f(x+5) \leqslant f(x+4)+1 \\
& \leqslant f(x+3)+2 \leqslant f(x+2)+3 \\
& \leqslant f(x+1)+4 \leqslant f(x)+5,
\end{aligned}
$$

Therefore, all the equalities hold, hence
$$
f(x+1)=f(x)+1 .
$$

Thus, from $f(1)=1$, we get $f(2)=2, f(3)=3, \cdots, f(2002)=$ 2002, so
$$
g(2002)=f(2002)+1-2002=1 .
$$