$14 \cdot 74$ For a real number $x$, $[x]$ denotes the greatest integer not exceeding $x$. Find the positive integer $n$ such that
$$
\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+\left[\log _{2} n\right]=1994
$$

holds.
(12th American Invitational Mathematics Examination, 1994)
［Solution］Let $S_{n}=\sum_{i=1}^{n}\left[\log _{2} i\right]$.
Notice that, for non-negative integer $k$, there are $2^{k}$ positive integers $x$ such that $\left[\log _{2} x\right]=k$, i.e., $x=$ $2^{k}, 2^{k}+1, \cdots, 2^{k+1}-1$.
Therefore, for positive integer $r$, we have
$$
\begin{aligned}
S_{2^{r}-1}= & 0+(1+1)+(2+2+2+2)+\cdots \\
& +\underbrace{[(r-1)+(r-1)+\cdots+(r-1)]}_{2^{r-1 \uparrow}} \\
= & 1 \cdot 2+2 \cdot 2^{2}+\cdots+(r-1) \cdot 2^{r-1} \\
= & (r-1) \cdot 2^{r}-\left(2^{r}-2\right) \\
= & (r-2) \cdot 2^{r}+2
\end{aligned}
$$

Let $r=8$, we get $S_{255}=15381994$.
Therefore, if $S_{n}=1994$, then $2^{8}-1<n<2^{9}-1$,
$$
\begin{aligned}
1994 & =S_{n}=S_{255}+(n-255) \cdot 8 \\
& =1538+8 n-255 \times 8=8 n-502 .
\end{aligned}
$$

So $n=312$.