Example 3. Let the sum of 10 distinct positive odd numbers be 1990, find the maximum value of the product of these 10 numbers.
Let the positive odd number be $2 x_{k}-1(k=1,2, \cdots$, 10 ), where all $x$ are unequal, then
$$
\sum_{k=1}^{10}\left(2 x_{k}-1\right)=1990,
$$

It is easy to know $\sum_{k=1}^{10} x_{k}=\frac{1}{2}(1990+10)=1000$.
Let $\varphi(x)=2 x-1$,
$$
\text { This problem is to find } F=\prod_{k=1}^{10} \varphi\left(x_{k}\right)=\prod_{k=1}^{10}\left(2 x_{k}-1\right)
$$

the maximum value of type $N(1000,1)$.
Since
$$
\begin{array}{l} 
\varphi(x-1) \varphi(x+1) \\
=(2 x-3)(2 x+1) \\
<(2 x-1)^{2}=[\varphi(x)]^{2},
\end{array}
$$

By Theorem 3 (Simplification $-\frac{1}{2} n(n-1)=955$, $q=95, \quad r=5$ ) we get
$$
\begin{aligned}
\pi_{\mathrm{max}} & =\prod_{\mathrm{b}=1}^{0} \varphi(94+k) \cdot \prod_{k=8}^{10} \varphi(95+k) \\
& =\prod_{k=1}^{\delta}(187+2 k) \cdot \prod_{k=0}^{10}(189+2 k) \\
& =189 \cdot 191 \cdots 197 \cdot 201 \cdot 203 \\
& \cdots 209 .
\end{aligned}
$$

Note 4. In Theorems $1-3$, if the variables are also restricted to $m \leqslant x_{1} \leqslant M(k=1,2, \cdots, n)$,
then when $m \leqslant q \leqslant M-(n-1) d-1$, the conclusions of each theorem give the maximum value of type $N(S, d)$ of $F$ within the range (9), and
at this time, it is only necessary for (4) to hold for $m+1 \leqslant a \leqslant b \leqslant M-1$, $m \leqslant c_{\mathrm{k}} \leqslant M$, or (7), (8) to hold for $m+1 \leqslant x \leqslant M-1$.