11. (20 points) Let there be $m$ distinct positive even numbers and $n$ distinct positive odd numbers whose sum is 2015. Find the maximum value of $20 m+15 n$.


11. From the problem, we know that \( n \) is an odd number.

And \( 20 m + 15 n = 5(4 m + 3 n) \), so we only need to find the maximum value of \( 4 m + 3 n \).

Let the \( m \) distinct positive even numbers and \( n \) distinct positive odd numbers, in ascending order, be
\[
\begin{array}{l}
a_{1} < a_{2} < \cdots < a_{m}, b_{1} < b_{2} < \cdots < b_{n}. \\
\text { Then } a_{1} + a_{2} + \cdots + a_{m} \\
\geqslant 2 + 4 + \cdots + 2 m = m(m + 1), \\
b_{1} + b_{2} + \cdots + b_{n} \\
\geqslant 1 + 3 + \cdots + (2 n - 1) = n^2.
\end{array}
\]

Thus, \( a_{1} + a_{2} + \cdots + a_{m} \)

Hence, \( m(m + 1) + n^2 \leqslant \sum_{i=1}^{m} a_{i} + \sum_{j=1}^{n} b_{j} = 2015 \)
\[
\Rightarrow (2 m + 1)^2 + (2 n)^2 \leqslant 2015 \times 4.
\]

Using the Cauchy-Schwarz inequality, we get
\[
\begin{array}{l}
{\left[(2 m + 1)^2 + (2 n)^2\right](16 + 9)} \\
\geqslant [3(2 m + 1) + 8 n]^2 \\
\Rightarrow 2015 \times 4 \times 25 \geqslant [4(2 m + 1) + 6 n]^2 \\
\Rightarrow 4 m + 3 n \leqslant 222.
\end{array}
\]

Since \( n \) is an odd number, \( 4 m + 3 n \) is also an odd number. Therefore,
\[
4 m + 3 n \leqslant 221.
\]

Finally, we attempt to construct a set of numbers
\[
a_{1}, a_{2}, \cdots, a_{m}, b_{1}, b_{2}, \cdots, b_{n},
\]
such that \( 4 m + 3 n = 221 \).
According to the condition for equality in the Cauchy-Schwarz inequality, we should choose suitable pairs of positive integers \((m, n)\) that satisfy
\[
\left\{\begin{array}{l}
4 m + 3 n = 221, \\
\frac{2 m + 1}{2 n} \approx \frac{4}{3}, \\
n \text { is an odd number. }
\end{array}\right.
\]

For this, we take \( m = 35 \) and \( n = 27 \), at this time,
\[
m(m + 1) + n^2 = 1989 < 2015.
\]

Thus, we take \( b_{i} = 2 i - 1 (i = 1, 2, \cdots, 27) \),
\[
\begin{array}{l}
a_{j} = 2 j (j = 1, 2, \cdots, 34), \\
a_{27} = 2 \times 35 + 26 = 96,
\end{array}
\]
which satisfies the conditions.
In summary, the maximum value that \( 20 m + 15 n \) can take is \( 5 \times 221 = 1105 \).