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. Let the indices $n$ that satisfy $S_{n}=0$ be arranged 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, without loss of generality, assume $S_{b_{k}}=0$.
Thus, by Table 1, it is easy to prove by mathematical induction:
$$
\left\{\begin{array}{l}
S_{b_{k}+2 i-1}=b_{k}+2-i, \\
S_{b_{k}+2 i}=2 b_{k}+i+2
\end{array}\left(i=1,2, \cdots, b_{k}+2\right)\right. \text {. }
$$

Table 1
\begin{tabular}{|c|c|c|}
\hline$n$ & $S_{n}$ & $a_{n}$ \\
\hline$b_{k}$ & 0 & $/$ \\
\hline$b_{k}+1$ & $b_{k}+1$ & $b_{k}+1$ \\
\hline$b_{k}+2$ & $2 b_{k}+3$ & $b_{k}+2$ \\
\hline$b_{k}+3$ & $b_{k}$ & $-\left(b_{k}+3\right)$ \\
\hline$b_{k}+4$ & $2 b_{k}+4$ & $b_{k}+4$ \\
\hline$b_{k}+5$ & $b_{k}-1$ & $-\left(b_{k}+5\right)$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ \\
\hline
\end{tabular}

Let $b_{k}+2-i=0$, we have
$$
b_{k}+2 i-1=3 b_{k}+3 \text {, }
$$

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

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