$164[x]$ represents the greatest integer not exceeding $x$, then $\sum_{k=1}^{99}[\sqrt{k}]=$
164615 . Obviously, when $i^{2} \leqslant k \leqslant(i+1)^{2}-1=i^{2}+2 \cdot i(i=1,2,3, \cdots, 9)$, we have $[\sqrt{k}]=i$, so
$$
\begin{aligned}
\sum_{k=1}^{99}[\sqrt{k}] & =\sum_{i=1}^{9} i(2 i+1)=\sum_{i=1}^{9} 2 i(i-1)+\sum_{i=1}^{9} 3 i \\
& =4 \sum_{i=2}^{9} C_{i}^{2}+\frac{3 \times 9 \times 10}{2} \\
& =4 \times C_{10}^{3}+135=615 .
\end{aligned}
$$