6. (1988 Mathematics Retest Question for Science Experimental Class Admission) Let $S=[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{1988}]$, find $[\sqrt{S}]$.

保留源文本的换行和格式，翻译结果如下：

6. (1988 Mathematics Retest Question for Science Experimental Class Admission) Let $S=[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{1988}]$, find $[\sqrt{S}]$.
6. Note that,

if $x \in M=\left\{n^{2}, n^{2}+1, \cdots,(n+1)^{2}-1\right\}$, then $[\sqrt{x}]=n$.
And $M$ has $2 n+1$ elements, thus
$$
\left[\sqrt{n^{2}}\right]+\left[\sqrt{n^{2}+1}\right]+\cdots+\left[\sqrt{(n+1)^{2}-1}\right]=n \cdot(2 n+1) \text {. }
$$

Since $44^{2}=1936$, we have
$$
[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{1935}]=1 \cdot 3+2 \cdot 5+\cdots+43 \cdot 87 \text {. }
$$

Also, $1988-1936+1=53$, so
$$
[\sqrt{1936}]+[\sqrt{1937}]+\cdots+[\sqrt{1988}]=44 \cdot 53 \text {. }
$$

Therefore,
$$
S=1 \cdot 3+2 \cdot 5+\cdots+43 \cdot 87+44 \cdot 53=58146 .
$$

Thus, $[\sqrt{S}]=241$.