Four, (Full marks 20 points) The sum of \( m \) different positive even numbers divisible by 5 and \( n \) different positive odd numbers divisible by 3 is \( M \). For all such \( m \) and \( n \), the maximum value of \( 5m + 3n \) is 123. What is the maximum value of \( M \)? Please prove your conclusion.

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The sum of \( m \) different positive even numbers divisible by 5 and \( n \) different positive odd numbers divisible by 3 is \( M \). For all such \( m \) and \( n \), the maximum value of \( 5m + 3n \) is 123. We need to find the maximum value of \( M \).

### Step-by-Step Solution:

1. **Identify the sequences:**
   - The sequence of positive even numbers divisible by 5 is: \( 10, 20, 30, 40, \ldots \)
   - The sequence of positive odd numbers divisible by 3 is: \( 3, 9, 15, 21, \ldots \)

2. **Sum of the sequences:**
   - The sum of the first \( m \) terms of the sequence \( 10, 20, 30, \ldots \) is:
     \[
     S_m = 10 + 20 + 30 + \cdots + 10m = 10(1 + 2 + 3 + \cdots + m) = 10 \cdot \frac{m(m+1)}{2} = 5m(m+1)
     \]
   - The sum of the first \( n \) terms of the sequence \( 3, 9, 15, \ldots \) is:
     \[
     S_n = 3 + 9 + 15 + \cdots + 3n = 3(1 + 3 + 5 + \cdots + (2n-1)) = 3 \cdot n^2
     \]

3. **Total sum \( M \):**
   \[
   M = 5m(m+1) + 3n^2
   \]

4. **Maximize \( 5m + 3n \):**
   - Given \( 5m + 3n = 123 \), we need to find the maximum value of \( M \).

5. **Express \( n \) in terms of \( m \):**
   \[
   n = \frac{123 - 5m}{3}
   \]
   - \( n \) must be a positive integer, so \( 123 - 5m \) must be divisible by 3.

6. **Check divisibility:**
   - \( 123 \equiv 0 \pmod{3} \)
   - \( 5m \equiv 0 \pmod{3} \) implies \( m \equiv 0 \pmod{3} \)
   - Let \( m = 3k \) for some integer \( k \).

7. **Substitute \( m = 3k \):**
   \[
   n = \frac{123 - 5(3k)}{3} = \frac{123 - 15k}{3} = 41 - 5k
   \]
   - \( n \) must be a positive integer, so \( 41 - 5k > 0 \)
   - \( k < \frac{41}{5} \approx 8.2 \)
   - The largest integer \( k \) is 8.

8. **Calculate \( m \) and \( n \):**
   - \( m = 3 \times 8 = 24 \)
   - \( n = 41 - 5 \times 8 = 1 \)

9. **Calculate \( M \):**
   \[
   M = 5m(m+1) + 3n^2 = 5 \cdot 24 \cdot 25 + 3 \cdot 1^2 = 3000 + 3 = 3003
   \]

### Conclusion:
The maximum value of \( M \) is \(\boxed{3003}\).
From the problem, we know
$$
\begin{aligned}
M \geqslant & 5(2+4+\cdots+2 m) \\
& +3[1+3+5+\cdots+(2 n-1)] \\
= & 5 m(m+1)+3 n^{2} .
\end{aligned}
$$

That is, \( 5\left(m^{2}+m\right)+3 n^{2} \leqslant M \),
$$
5\left(m+\frac{1}{2}\right)^{2}+3 n^{2} \leqslant M+\frac{5}{4} \text {. }
$$

By the Cauchy-Schwarz inequality, we know
$$
\begin{array}{l}
5\left(m+\frac{1}{2}\right)+3 n \\
\leqslant \sqrt{(\sqrt{5})^{2}+(\sqrt{3})^{2}} \\
\quad \sqrt{\left[\sqrt{5}\left(m+\frac{1}{2}\right)\right]^{2}+(\sqrt{3} n)^{2}} \\
\leqslant \sqrt{8} \cdot \sqrt{M+\frac{5}{4}}=\sqrt{8 M+10}
\end{array}
$$

Thus, \( 5 m+3 n \leqslant \sqrt{8 M+10}-\frac{5}{2} \).
Since the maximum value of \( 5 m+3 n \) is 123,
$$
\begin{array}{l}
\therefore 123 \leqslant \sqrt{8 M+10}-\frac{5}{2}<124 . \\
\therefore 1967.5312 \leqslant M<1999.0312 .
\end{array}
$$
Thus, the maximum value of \( M \) does not exceed 1999.
When \( 5 m+3 n=123 \),
$$
\begin{array}{l}
m=3 k, n=41-5 k(k \in \mathbf{N}) . \\
\because 5 m^{2} \leqslant M<2000,3 n^{2} \leqslant M<2000, \\
\therefore m<20, n<26 . \\
\therefore 3 k<20,41-5 k<26 .
\end{array}
$$

Solving this, we get \( 3<k<\frac{20}{3} \).
Thus, \( k \) is 4, 5, 6.
Then the three pairs of values for \( (m, n) \) are
\( (12,21) \cdot(15,16) \cdot(18,11) \).
It is easy to verify that the pairs \( (12,21) \) and \( (18,11) \) do not satisfy the inequality
$$
5 m(m+1)+3 n^{2} \leqslant M<2000 .
$$

Only \( (15,16) \) satisfies:
$$
5 \times 15(15+1)+3 \times 16^{2}=1968<2000 \text {. }
$$

And it is possible to select 15 positive even numbers divisible by 5 and 16 positive odd numbers divisible by 3 such that their sum is 1998. For example:
$$
\begin{array}{l}
5(2+4+\cdots+28+30)+3(1+3+\cdots+29+31) \\
=1968, \\
5(2+4+\cdots+28+30)+3(1+3+\cdots+29+41) \\
=1998 .
\end{array}
$$

There do not exist integers \( m \) and \( n \) such that \( M \) is 1999. Otherwise, we would have
$$
5 \times 2 m_{0}+3 \times 2 n_{0}=1999-1968=31 \text {. }
$$

This equation has no integer solutions.
Therefore, the maximum value of \( M \) is 1998.