$$
14 \cdot 32 \text { Let } S=[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{1988}] \text {. Find }[\sqrt{S}] \text {. }
$$
(Advanced Class for Science Experiment Mathematics Admission Test, 1988)
［Solution］Notice that,
if $x \in M=\left\{n^{2}, n^{2}+1, \cdots,(n+1)^{2}-1\right\}$, then
$$
[\sqrt{x}]=n \text {. }
$$

$M$ contains $2 n+1$ elements. Therefore,
$$
\begin{array}{l}
\quad\left[\sqrt{n^{2}}\right]+\left[\sqrt{n^{2}+1}\right]+\cdots+\left[\sqrt{(n+1)^{2}-1}\right] \\
=n \cdot(2 n+1) .
\end{array}
$$

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

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

Thus,
$$
\begin{aligned}
S & =1 \cdot 3+2 \cdot 5+\cdots+43 \cdot 87+44 \cdot 53 \\
& =58146 .
\end{aligned}
$$

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