13. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+\cdots+b_{n} x^{n},
\end{array}
$$

and $b_{1}+b_{2}+\cdots+b_{n}=1013$.
Then a possible value of the positive integer $n$ is
13. 9 .

Let $x=0$, we get $b_{0}=n$.
Let $x=1$, we get
$$
2+2^{2}+\cdots+2^{n}=b_{0}+b_{1}+\cdots+b_{n},
$$

which is $2\left(2^{n}-1\right)=n+1013$.
Solving this, we get $n=9$.