In $1,2, \cdots, 100$ these 100 positive integers, remove 50 so that in the remaining positive integers, any two different $a, b$ have $a \nmid b$. Find the maximum possible value of the sum of all removed positive integers.
Let the remaining numbers be $a_{i}=2^{3} t_{i}$, where $i \in \{1,2, \cdots, 50\}$, $s_{i}$ is a natural number, and $t_{i}$ is an odd number.

Since for any $1 \leqslant i \neq j \leqslant 50$, we have $a_{i} \nmid a_{j}$, it follows that $t_{i} \neq t_{j}$, meaning $t_{1}, t_{2}, \cdots, t_{50}$ are 50 distinct odd numbers.

The numbers $1,2, \cdots, 100$ can be expressed in the form $2^{s} t$, where $t$ has only 50 possibilities, namely $1,3, \cdots, 99$. Therefore,
$$
\left\{t_{1}, t_{2}, \cdots, t_{50}\right\}=\{1,3, \cdots, 99\}.
$$

Without loss of generality, let $t_{i}=2 i-1$.
For convenience, we can change the indices of the remaining numbers.
Let $a_{i}=2^{s} i$, where $i \in \{1,3, \cdots, 99\}$ and $s_{i}$ is a natural number.
Set $S_{0}=\left\{a_{99}, a_{97}, \cdots, a_{35}\right\}$,
$S_{1}=\left\{a_{33}, a_{31}, \cdots, a_{13}\right\}$,
$S_{2}=\left\{a_{11}, \cdots, a_{5}\right\}$, $S_{3}=\left\{a_{3}\right\}$, $S_{4}=\left\{a_{1}\right\}$.
First, we prove that for any $a_{i} \in S_{k}$, we have
$a_{i} \geqslant 2^{k} i$.
We use mathematical induction on $k$.
When $k=0$, $a_{i}=2^{s} t \geqslant 2^{0} i$ holds.
Assume that for $k=m (0 \leqslant m \leqslant 3)$, the conclusion holds.
Then for $k=m+1$, if $a_{i}=2^{s} i \in S_{m+1}$, we have $a_{3 i} \in S_{m}$.
Since $a_{i} \nmid a_{3 i}$, by the induction hypothesis,
$a_{3 i}=2^{s_{3 i}} 3 i (s_{3 i} \geqslant m)$.
Thus, $s_{i} \geqslant s_{3 i}+1 \geqslant m+1$, which means $a_{i} \geqslant 2^{m+1} i$, and the conclusion holds.
By mathematical induction, equation (1) is proven.
Next, we prove that equation (1) holds with equality when the conditions of the problem are satisfied.
If $a_{i}, a_{j} \in S_{k}$, then $a_{i}=2^{k} i, a_{j}=2^{k} j$.
Assume $a_{i} \mid a_{j}$. Then $i \mid j$.
Since $i, j$ are both odd, it follows that $j \geqslant 3 i$.
By $a_{i}, a_{j} \in S_{k}$, we know $3 i > j$, which is a contradiction. Hence, $a_{i} \nmid a_{j}$.
Similarly, $a_{j} \nmid a_{i}$.
If $a_{i} \in S_{k}, a_{j} \in S_{l}$, without loss of generality, let $l > k$, then $a_{i}=2^{k} i, a_{j}=2^{l} j$.
Since $l > k$, it follows that $a_{i} \nmid a_{j}$.
Assume $a_{j} \mid a_{i}$. Then $j \mid i$.
By $l > k$, we know $j > i$, which is a contradiction. Hence, $a_{j} \nmid a_{i}$.
In summary, the sum of the remaining numbers is minimized when $a_{i}=2^{k} i$.
Thus, the minimum value is
$2211 + 506 + 128 + 24 + 16 = 2885$.
Therefore, the maximum possible value of the sum of all removed positive integers is $5050 - 2885 = 2165$.