Example. Given $f(x)=[3 x]+[4 x]+[12 x]+[16 x]$.
(1) Find the number of positive integer values $M$ that $f(x)$ takes when $0<x \leqslant 1$.
(2) Is 97 a value of $f(x)$?
$$
\begin{array}{l}
\text { (1) } \sum n_{i}=3+4+12+16=35 . \\
\sum_{1<i<j<1}\left(n_{i}, n_{j}\right)=(3,4)+(3,12)+(3,16)+(4,12)+(4,16)+(12,16)=17. \\
\sum_{1<i<j<k<t}\left(n_{i}, n_{j}, n_{k}\right)=(3,4,12)+(3,4,16)+(3,12,16)+(4,12,16)=7. \\
\left(n_{i}, n_{2}, n_{3}, n_{!}\right)=(3,4,12,16)=1, \\
\end{array}
$$

By Proposition 1, \( M=35-17+7-1=24 \).
$$
\begin{array}{l}
\text { (2) } \because N=9 \cdot \because=2 \times 35+27, r=27, \\
\text { } D=[3,4,12,16]=48, \\
\therefore \frac{D}{\sum_{i=1}^{4} n_{i}} \cdot r=\frac{48}{35} \cdot 27=36.99, \\
\frac{D}{\sum_{i=1}^{4} n_{i}}(r+4-1)=\frac{48}{35} \cdot 30=41.1 .
\end{array}
$$

From \( 36.99 \leqslant t^{\prime}<41.1 \), we know the possible choices are: \( 37,38,39,40,41 \), but only 39 and 40 are multiples of \( \frac{48}{3}=16, \frac{48}{4}=12, \frac{48}{12}=4, \frac{48}{16}=3 \), so we discard \( 37,38,41 \).
\(\because\) Substituting \( x=\frac{39}{48} \) we get:
$$
\begin{array}{l}
f\left(\frac{39}{48}\right)=2+3+9+13=27, \\
\begin{aligned}
\therefore f\left(2 \frac{39}{48}\right)= & f(2)+f\left(\frac{39}{48}\right) \\
= & 2 \times 35+27=97,
\end{aligned}
\end{array}
$$

Thus, 97 is a value of \( f(x)=[3 x]+[4 x]+[12 x]+[16 x] \).