Suppose $n = 2^k$, where $k$ is a positive integer. We will prove by mathematical induction that from any set containing $2n - 1$ positive integers, one can always select $n$ integers such that the sum of these $n$ integers is divisible by $n$. When $k = 10$, this is exactly what the problem asks.

1. When $k=1$, $n=2$, $2n - 1 = 3$. Among any three positive integers, there must be two numbers with the same parity, and their sum is divisible by 2.

2. Suppose the statement holds when $n = 2^k$. Consider $n = 2^{k+1}$. Since $2n - 1 = 2^{k+2} - 1 > 2^{k+1} - 1$, by the induction hypothesis, there exist $2^k$ positive integers such that their sum is divisible by $2^k$. Furthermore, $2^{k+2} - 1 - 2^k = 3 \times 2^k - 1 > 2^{k+1} - 1$, so by the induction hypothesis, there exist $2^k$ positive integers such that their sum is divisible by $2^k$. Also, $3 \times 2^k - 1 - 2^k = 2^{k+1} - 1$, so by the induction hypothesis, there exist $2^k$ positive integers such that their sum is divisible by $2^k$. Denote the sums of these three groups of positive integers by $2^k a, 2^k b, 2^k c$ respectively. Among these three numbers $a, b, c$, at least two must have the same parity; without loss of generality, suppose $a + b = 2m$. Then these two groups of numbers together consist of $2^k + 2^k = 2^{k+1}$ numbers, and their sum is
$$2^k a + 2^k b = 2^k (a + b) = 2^k \times 2m = 2^{k+1}m,$$
which is divisible by $2^{k+1}$.