$3 \cdot 66$ What is the largest even number that cannot be expressed as the sum of two odd composite numbers?
(2nd American Invitational Mathematics Examination, 1984)
[Solution] We prove: For even numbers not less than 40, they can definitely be expressed as the sum of two odd composite numbers.

We notice that if $n$ is an odd number greater than 1, then $5n$ is an odd composite number with a units digit of 5.
To express all even numbers as the sum of two odd composite numbers, one odd composite number can be $5n$,
and the other odd composite number only needs to be an odd composite number with a units digit of 5, 7, 9, 1, or 3. It is easy to see that 15, 27, 9, 21, and 33 are odd composite numbers that meet the requirements.
Let $k$ be an even number not less than 40.
If the units digit of $k$ is 2 (i.e., $k=42,52,62, \cdots$), then
$$
k=27+5n.
$$

If the units digit of $k$ is 4 (i.e., $k=44,54,64, \cdots$), then
$$
k=9+5n.
$$

If the units digit of $k$ is 6 (i.e., $k=46,56,66, \cdots$), then
$$
k=21+5n.
$$

If the units digit of $k$ is 8 (i.e., $k=48,58,68, \cdots$), then
$$
k=33+5n.
$$

If the units digit of $k$ is 0 (i.e., $k=40,50,60, \cdots$), then
$$
k=15+5n.
$$

Thus, even numbers that cannot be written as the sum of two odd composite numbers must be less than 40.
The largest even number is 38.
Since $38=1+37=3+35=5+33=7+31=9+29=11+27$ $=13+25=15+23=17+21=19+19$, 38 cannot be expressed as the sum of two odd composite numbers.
Therefore, 38 is the even number that meets the requirements of the problem.