Given a height of 241, find the minimum value of the function
$$
S_{n}=|n-1|+2|n-2|+\cdots+100|n-100|
$$
$\left(n \in \mathbf{N}_{+}\right)$.
Solution: When $n \geqslant 100$,
$$
\begin{array}{l}
S_{n}=(n-1)+2(n-2)+\cdots+100(n-100) \\
=n(1+2+\cdots+100)-\left(1^{2}+2^{2}+\cdots+10^{2}\right) \\
\geqslant S_{100} .
\end{array}
$$

Therefore, the minimum value of $S_{n}$ can only be obtained when $1 \leqslant n \leqslant 100$.
When $2 \leqslant n \leqslant 100$,
$$
\begin{array}{l} 
S_{n}-S_{n-1} \\
=|n-1|+|n-2|+\cdots+ \\
|n-100|-100|n-101| \\
=|n-1|+|n-2|+\cdots+ \\
|n-100|-100(101-n) \\
=|n-1|+|n-2|+\cdots+ \\
|n-100|-100 \times 101+100 n .
\end{array}
$$

Let the above expression be $f(n)$.
Thus, when $3 \leqslant n \leqslant 100$,
$$
\begin{array}{l}
f(n)-f(n-1)=2 n-2>0 . \\
\text { Hence, } f(2) < 0 .
\end{array}
$$

Notice that
$$
\begin{array}{c}
f(71)=70+69+\cdots+1+0+1+ \\
2+\cdots+29-100 \times 30 \\
=-800 .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } f(2) < 0, \text{ and } S_{2} > S_{3} > \cdots > S_{70} > S_{71}, \\
S_{71} < S_{72} < \cdots < S_{99} < S_{100} .
\end{array}
$$

In summary, the minimum value of $S_{n}$ is obtained if and only if $n=71$.
Now, let's find the minimum value.
$$
\begin{array}{l}
S_{71}=|71-1|+2|71-2|+\cdots+ \\
71|71-71|+72|71-72|+\cdots+ \\
100|71-100| \\
=(1+2+\cdots+70) \times 71- \\
\left(1^{2}+2^{2}+\cdots+70^{2}\right)+ \\
\left(72^{2}+73^{2}+\cdots+100^{2}\right)- \\
71(72+73+\cdots+100) \\
=\frac{70 \times 71}{2} \times 71-\frac{70 \times 71 \times 141}{6} \times 2-71^{2}+ \\
\frac{100 \times 101 \times 201}{6}-\frac{172 \times 29}{2} \times 71 \\
=176435-233590-5041+ \\
338350 - 177074 \\
=99080 .
\end{array}
$$