Example 5 Let $x_{1}, x_{2}, \cdots, x_{100}$ be positive integers, and $x_{1}<x_{2}<\cdots<x_{100}$. If $x_{1}+x_{2}+\cdots+x_{100}=7001$, then the maximum value of $x_{1}+x_{2}+\cdots+x_{50}$ is ( ).
(A) 2225
(B) 2226
(C) 2227
(D) 2228
Because
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{50} \\
\leqslant 50 x_{50}-(1+2+\cdots+49) \\
=50 x_{50}-1225,
\end{array}
$$

Therefore, we need to determine the value of $x_{50}$.
To maximize the sum $x_{1}+x_{2}+\cdots+x_{50}$, the latter part should be minimized.
$$
\begin{array}{l}
\text { Hence } x_{51}+x_{52}+\cdots+x_{100} \\
\geqslant 50 x_{50}+(1+2+\cdots+50) . \\
=50 x_{50}+1275 .
\end{array}
$$

By taking the maximum and minimum values of the above two expressions, we get
$$
\begin{array}{l}
\left(x_{1}+x_{2}+\cdots+x_{50}\right)+\left(x_{51}+x_{52}+\cdots+x_{100}\right) \\
=100 x_{50}+50=7001 .
\end{array}
$$

Therefore, $x_{30}=69.51$.
Since $x_{50}$ is an integer, we consider the following two scenarios.
(1) If $x_{50}=70$, then $x_{1}, x_{2}, \cdots, x_{100}$ are 21, $22, \cdots, 120$, and their sum 7050 is 49 more than 7001. Only by subtracting 1 from each of the first 49 numbers can we achieve this, so the maximum value of the first 50 numbers is
$$
50 \times 70-1225-49=2226 \text {; }
$$
(2) If $x_{50}=69$, then $x_{1}, x_{2}, \cdots, x_{100}$ are 20, $21, \cdots, 119$, and their sum 6950 is 51 less than 7001. Only by adding 1 to each of the last 51 numbers can we achieve this, so the maximum value of the first 50 numbers is
$$
50 \times 69-1225+1=2226 .
$$