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

If $g(x)=f(x)+1-x$, then $g(2002)=$
10.1.
$$
\begin{array}{l}
\text { From } g(x)=f(x)+1-x \text { we get } f(x)=g(x)+x-1 . \\
\text { Then } g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \text {, } \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 . \\
\text { Therefore, } g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \\
\therefore g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \\
\quad \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x) . \\
\therefore g(x+1)=g(x) .
\end{array}
$$

Thus, $g(x)$ is a periodic function with a period of 1.
Given $g(1)=1$, hence $g(2002)=1$.