3. Let $k$ be a given positive integer, find the smallest positive integer $n$, such that among any $n$ integers, there always exist two positive integers, the sum or difference of which is divisible by $2k$.


3. Take the $k+1$ integers $0,1,2, \cdots, k$. The sum of any two of these numbers lies between 1 and $2 k-1$, and the absolute value of their difference lies between 1 and $k-1$, none of which can be divisible by $2 k$. Therefore, the smallest positive integer required is $\geqslant k+2$. On the other hand, divide the remainders modulo $2 k$ into $k+1$ groups: $\{0\},\{1,2 k-1\},\{2,2 k-2\}, \cdots,\{k-1, k+1\},\{k\}$. For any $k+2$ integers, at least two of their remainders when divided by $2 k$ belong to the same group, and the sum of two remainders in the same group (which can be the same) is $2 k$ or their difference is 0. Therefore, any $k+2$ integers must have two numbers whose sum or difference is divisible by $2 k$. In summary, the smallest value of $n$ is $k+2$.