Let $n$ be supereven and $d$ its greatest odd divisor. If $\frac{n}{d}$ were divisible by some odd $p > 1$, then $pd > d$ would also be a factor of $n$, contradiction. Thus $\frac{n}{d}$ is a power of 2. Hence supereven numbers are exactly those that can be expressed as a product of a power of two and an odd number less than that power of two.
We will find the supereven numbers less than 1000 by their largest odd factors $d$.

* If $d = 1$, then the power of 2 can be one of 2, 4, 8, 16, 32, 64, 128, 256, 512. We obtain 9 supereven numbers.

* If $d = 3$, then the power of 2 can be one of 4, 8, 16, 32, 64, 128, 256. We obtain 7 supereven numbers.

* If $d = 5$ or $d = 7$, then the power of 2 can be one of 8, 16, 32, 64, 128. We obtain $2 \cdot 5 = 10$ supereven numbers.

* If $d$ is one of the numbers 9, 11, 13, 15, then the power of 2 can be one of 16, 32, 64. We obtain $4 \cdot 3 = 12$ supereven numbers.

* If $d$ is one of the numbers 17, 19, 21, 23, 25, 27, 29, 31, then the power of 2 can be only 32. We obtain 8 supereven numbers.

Thus there are $9 + 7 + 10 + 12 + 8 = 46$ supereven numbers less than 1000.

Like in Solution 1, we will show that supereven numbers are exactly those that can be expressed as a product of a power of two and an odd number less than that power of 2. We will call this power of 2 the *even part* of a supereven integer. As every other integer is odd, there are $\frac{2^k}{2} = 2^{k-1}$ odd positive integers less than $2^k$. The exponents $k = 1, 2, 3, 4, 5$ thus yield 1, 2, 4, 8, 16 supereven integers with an even part of $2^k$. Of those, the greatest is $2^5 \cdot (2^5 - 1) = 992 < 1000$, so all of them are less than 1000. For $k = 6, 7, 8, 9$, the greatest odd factor keeping the product under 1000 is 15, 7, 3, 1 respectively, yielding $8 + 4 + 2 + 1 = 15$ more supereven numbers less than 1000. Thus the total number of supereven integers less than 1000 is $1 + 2 + 4 + 8 + 16 + 15 = 46$.