42 Let $k$ be a natural number. Try to determine the smallest natural number $n$ such that: in any $n$ integers, there must be two numbers whose sum or difference is divisible by $2 k+1$.


42 Consider the set $\{k+1, k+2, \cdots, 2 k+1\}$ containing $k+1$ numbers. The sum of any two different numbers lies between $2 k+3$ and $4 k+1$, and thus cannot be divisible by $2 k+1$. The difference between any two numbers lies between 1 and $k$, and thus cannot be divisible by $2 k+1$ either. Therefore, $n \geqslant k+2$.
On the other hand, let the set $\Lambda$ have $k+2$ integers. If there are two numbers in set $A$ that are congruent modulo $2 k+1$, then their difference is divisible by $2 k+1$, and $A$ meets the requirement. Otherwise, the remainders of the numbers in set $A$ modulo $2 k+1$ are all different. We can assume that the set of remainders is a $(k+2)$-element subset of the set
$$
\begin{aligned}
M & =\{-k,-(k-1), \cdots,-1,0,1, \cdots, k\} \\
& =\{-1,1\} \cup\{-2,2\} \cup \cdots \cup\{-k, k\} \cup\{0\} .
\end{aligned}
$$

Selecting any $k+2$ numbers from $M$, there must be two numbers in the same $\{j,-j\}$, whose sum is 0, so $A$ must have two numbers whose sum is divisible by $2 k+1$.
Thus, the minimum value of $n$ is $k+2$.