13. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Let
$$
\begin{array}{l}
S=[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{2019}] . \\
\text { Then }[\sqrt{S}]=
\end{array}
$$
13.243.

Let $k^{2} \leqslant n<(k+1)^{2}$. Then
$k \leqslant \sqrt{n}<k+1 \Rightarrow \sqrt{n}=k$.
Also, $(k+1)^{2}-k^{2}=2 k+1$, so from $k^{2} \sim(k+1)^{2}$ there are $2 k+1$ numbers, and their integer parts of the square roots are all $k$.
$$
\begin{array}{l}
\text { Hence }\left[\sqrt{k^{2}}\right]+\left[\sqrt{k^{2}+1}\right]+\cdots+\left[\sqrt{\left(k^{2}+1\right)^{2}-1}\right] \\
=k(2 k+1)=2 k^{2}+k .
\end{array}
$$

Also, by $44^{2}<2019=44^{2}+83<45^{2}$, there are 44 perfect squares between $1 \sim 2019$: $11^{2}, 2^{2}, \cdots, 44^{2}$.
$$
\begin{array}{l} 
\text { Hence } S=[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{2019}] \\
= 2\left(1^{2}+2^{2}+\cdots+43^{2}\right)+ \\
(1+2+\cdots+43)+44 \times 84 \\
= 59510 .
\end{array}
$$

And $243^{2}=59049,244^{2}=59536$, therefore,
$$
[\sqrt{S}]=[\sqrt{59510}]=243 \text {. }
$$