10. (20 points) In the sequence $\left\{a_{n}\right\}$, let $S_{n}=\sum_{i=1}^{n} a_{i}$ $\left(n \in \mathbf{Z}_{+}\right)$, with the convention: $S_{0}=0$. It is known that
$$
a_{k}=\left\{\begin{array}{ll}
k, & S_{k-1}<k ; \\
-k, & S_{k-1} \geqslant k
\end{array}\left(1 \leqslant k \leqslant n, k, n \in \mathbf{Z}_{+}\right)\right. \text {. }
$$

Find the largest positive integer $n$ not exceeding 2019 such that
$$
S_{n}=0 .
$$
10. Arrange the indices $n$ that satisfy $S_{n}=0$ in ascending order, denoted as the sequence $\left\{b_{n}\right\}$, then $b_{1}=0$.
To find the recurrence relation that $\left\{b_{n}\right\}$ should satisfy.
In fact, assume $S_{b_{k}}=0$.
Thus, by Table 1, it is easy to prove by mathematical induction:
$$
\begin{array}{l}
\left\{\begin{array}{l}
S_{b_{k}+2 i-1}=b_{k}+2-i, \\
S_{b_{k}+2 i}=2 b_{k}+i+2
\end{array}\left(i=1,2, \cdots, b_{k}+2\right) .\right. \\
\text { Let } b_{k}+2-i=0, \text { then } \\
b_{k}+2 i-1=3 b_{k}+3,
\end{array}
$$

Thus, the sequence $\left\{b_{n}\right\}$ satisfies:
$$
\begin{array}{l}
b_{k+1}=3 b_{k}+3, b_{1}=0 \\
\Rightarrow b_{n}=\frac{3^{n}-3}{2} .
\end{array}
$$

Therefore, $b_{7}=1092$ is the desired result.