China Team Selection Test

Let $S$ be a set of positive integers such that $n \in S$ if and only if
$$\sum_{d \mid n, d < n, d \in S} d \leq n.$$
We need to find all positive integers $n = 2^k \cdot p$ where $k$ is a non-negative integer and $p$ is an odd prime, such that
$$\sum_{d \mid n, d < n, d \in S} d = n.$$

Define the function
$$f(n) = \left( \sum_{d \mid n, d < n, d \in S} d \right) - n.$$
The given condition translates to $n \in S \Leftrightarrow f(n) \leq 0$. Additionally, we have $f(n) \leq \sigma(n) - 2n$, where $\sigma(n)$ denotes the sum of all positive divisors of $n$.

First, consider the powers of 2. For any $m \geq 0$,
$$f(2^m) \leq \sigma(2^m) - 2^{m+1} = -1,$$
which implies $2^m \in S$.

Next, observe that $1 \in S$ and for any odd prime $p$,
$$f(p) = 1 - 2p < 0,$$
thus $p \in S$.

Now, let $a = \lfloor \log_2(p+1) \rfloor - 1$. We claim that for $r \in \mathbb{N}$, $2^r p \notin S$ if and only if $a+1 \mid r$.

Proof of Claim:

Write $r = x(a+1) - y$ where $0 \leq y \leq a$ and $x \in \mathbb{N}$. We proceed by induction on $x \geq 1$.

Base Case:

For $x = 1$ and $y > 0$, we have $r \leq a$. Then,
$$f(2^r p) \leq \sigma(2^r p) - 2^{r+1} p = 2^{r+1} - p - 1 \leq 2^{a+1} - (p + 1) \leq 0,$$
so $2^r p \in S$ for $r \leq a$.

For $x = 1$ and $y = 0$,
$$f(2^r p) = f(2^{a+1} p) = \sigma(2^{a+1} p) - 2^{a+2} p = 2^{a+2} - (p + 1) > 0,$$
which means $2^{a+1} p \notin S$. This proves the base case.

Inductive Step:

Suppose the result is true for some $x$. Let $r = (x+1)(a+1) - y$. If $y > 0$, then $r \leq (x+1)(a+1) - 1$. Using the inductive hypothesis,
$$\begin{align*} f(2^r p) &\leq \sigma(2^r p) - 2^{r+1} p - \left( 2^{a+1} p + 2^{2(a+1)} p + \dots + 2^{x(a+1)} p \right) \\ &= 2^{r+1} - 1 - p - \frac{2^{a+1} (2^{x(a+1)} - 1) p}{2^{a+1} - 1} \\ &\leq (2^{(x+1)(a+1)} - 1) - \frac{(2^{(x+1)(a+1)} - 1) p}{2^{a+1} - 1} \\ &= (2^{(x+1)(a+1)} - 1) \left( 1 - \frac{p}{2^{a+1} - 1} \right) \leq 0, \end{align*}$$
so $2^r p \in S$ when $y > 0$.

If $y = 0$, then
$$\begin{align*} f(2^r p) &= \sigma(2^r p) - 2^{r+1} p - \left( 2^{a+1} p + 2^{2(a+1)} p + \dots + 2^{x(a+1)} p \right) \\ &= 2^{r+1} - 1 - p - \frac{2^{a+1} (2^{x(a+1)} - 1) p}{2^{a+1} - 1} \\ &= (2^{r+1} - 1) - \frac{(2^{(x+1)(a+1)} - 1) p}{2^{a+1} - 1} \\ &\geq (2^{(x+1)(a+1) + 1} - 1) - 2(2^{(x+1)(a+1)} - 1) = 1, \end{align*}$$
thus $2^{(x+1)(a+1)} p \notin S$, completing the induction step.

Conclusion:

The problem asks us to find all $k$ and $p$ such that $f(2^k p) = 0$. This implies equality must hold at all $\leq$ signs in the proof of the lemma above. The last equality holds only if $p = 2^{a+1} - 1$. Also, the second equality holds when $y = 1$, i.e., $r + 1 = x(a+1)$ for some positive integer $x$.

Therefore, all solutions are of the form $k = x(a+1) - 1$ and $p = 2^{a+1} - 1$ for $a, x \in \mathbb{N}$.

The answer is: $\boxed{k}$ = x(a+1) - 1 $\text{}$ and } p = $2^{a+1}$ - 1 $\text{}$ for } a, x ${\in \mathbb{N}}.$