26. Let $S=\{1,2,3,4, \ldots, 100000\}$. Find the least possible value of $k$ such that any subset $A$ of $S$ with $|A|=2010$ contains two distinct numbers $a$ and $b$ with $|a-b| \leq k$.
26. Answer: 49
Consider the following subsets of $S$ :
$$
\begin{array}{l} 
S_{1}=\{1,2,3, \ldots, 49,50\}, \\
S_{2}=\{1,52,53, \ldots, 99,100\}, \\
S_{3}=\{101,102,103, \ldots, 149,150\}, \\
\vdots \\
S_{2000}=\{99951,99952,99953, \ldots, 99999,100000\} .
\end{array}
$$

In other words, $S_{i}=\{50 i-49,50 i-48, \ldots, 50 i\}$ for $i=1,2, \ldots, 2000$. Note that $S$ is partitioned into these subsets $S_{1}, S_{2}, S_{3}, \ldots, S_{2000}$.
By Pigeonhole Principle, for any subset $A$ of $S$ with $|A|=2010$, there exists $i$, where $1 \leq i \leq 2000$, such that $\left|A \cap S_{i}\right| \geq 2$.
Let $a, b \in A \cap S_{i}$. It is clear that $|a-b| \leq 49$.
To show that 49 is the least possible value of $k$, we find a subset $A \subseteq S$ with $L A I=$ 2010 such that $|a-b| \geq 49$ for any distinct $a, b \in A$. Let
$$
A=\{49 j+1: j=0,1,2, \ldots, 2009\}=\{1,50,99,148, \ldots, 98442\} .
$$

Then $A$ is a subset of $S$ with $|A|=2010$ and $|a-b| \geq 49$ for any distinct $a, b \in A$.