3. The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 117. For all such $m$ and $n$, the maximum value of $3m + 2n$ is $\qquad$ .
3.37.

Let the $m$ positive even numbers be $a_{i}$, then
$$
\sum_{i=1}^{m} a_{i} \geqslant 2+4+\cdots+2 m=m(m+1) \text {. }
$$

Let the $n$ positive odd numbers be $b_{j}$, then
$$
\sum_{j=1}^{n} b_{j} \geqslant 1+3+\cdots+(2 n-1)=n^{2} .
$$

Thus, the problem reduces to finding the maximum value of $y=3 m+2 n$ under the condition $m(m+1)+n^{2} \leqslant 117$.
Let $m=-\frac{1}{2}+k \cos \theta, n=k \sin \theta$, then
$$
\begin{aligned}
k^{2} & =\left(m+\frac{1}{2}\right)^{2}+n^{2} \\
& =m(m+1)+n^{2}+\frac{1}{4} \leqslant 117+\frac{1}{4} .
\end{aligned}
$$

Thus,
$$
\begin{array}{l}
3 m+2 n=k(3 \cos \theta+2 \sin \theta)-\frac{3}{2} \\
\leqslant \sqrt{13} \cdot k \cdot \sin (\varphi+\theta)-\frac{3}{2} \\
\leqslant \sqrt{13} \times \sqrt{117+\frac{1}{4}}-\frac{3}{2}<37.6 .
\end{array}
$$

Therefore, the maximum value of $3 m+2 n$ is 37.
Since $m$ is an odd number,
by trial, when $m=9, n=5$, $9 \times 10+5^{2}<117$.