10. (20 points) There are 100 people standing in a circle on the playground, labeled as $C_{1}, C_{2}, \cdots, C_{100}$ in a clockwise direction. The host distributes souvenirs numbered $1,2, \cdots, 50$ to the people in the following manner: First, the souvenir numbered 1 is given to $C_{1}$; then, skipping 1 person in a clockwise direction, the souvenir numbered 2 is given to $C_{3}$; next, skipping 2 people in a clockwise direction, the souvenir numbered 3 is given to $C_{6}, \cdots \cdots$ On the $k$-th turn, $k$ people are skipped in a clockwise direction, and the souvenir numbered $k+1$ is given to $C_{\varphi(k+1)}$, where $\varphi(k) \equiv 1+2+\cdots+k(\bmod 100)$. This continues until all the souvenirs are distributed. Determine the person who receives the most souvenirs and the numbers of the souvenirs they receive.
10. Let the $s$-th and $t$-th (where $1 \leqslant s < t \leqslant 50$) souvenirs be obtained by the same person. Then
$$
100 \left\lvert\,\left[\frac{t(t+1)}{2}-\frac{s(s+1)}{2}\right]\right.,
$$

which is $200 I (t-s)(t+s+1)$.
Clearly, $t+s+1 \leqslant 100$, and $t-s$ and $t+s+1$ have different parities.
(1) If $81(t+s+1)$, then $25 \times(t+s+1)$.
(i) If $51(t+s+1)$, then
$$
t+s+1=40 k(k=1,2) \text {. }
$$

Also, $51(t-s)$, so $t \equiv s \equiv 2(\bmod 5)$.
Thus, $(s, t)=(5 r+2,40 k-5 r-3)$, where
$r=0,1, \cdots, 4 k-1 ; k=1,2$.
After calculation,
$$
\begin{aligned}
(s, t)= & (2,37),(7,32),(12,27), \\
& (17,22),(32,47),(37,42) .
\end{aligned}
$$
(ii) If $5 \times(t+s+1)$, then
$$
251(t-s) \Rightarrow t-s=25 \text {. }
$$
$$
\begin{array}{l}
\text { By } 8 I(t+s+1) \Rightarrow 8 \mid(2 s+26) \\
\Rightarrow 4 \mid(s+1) \Rightarrow s \equiv 3(\bmod 4) . \\
\text { Hence }(s, t)=(3,28),(7,32),(11,36), \\
(15,40),(19,44),(23,48) .
\end{array}
$$
(2) If $81(t-s)$, then $25 \times(t-s)$.
(i) If $51(t-s)$, then
$$
\begin{array}{l}
40 \mid(t-s) \Rightarrow t-s=40 . \\
\text { Also } 5|(t+s+1) \Rightarrow 5|(2 s+41) \\
\Rightarrow 5 ।(2 s+1) \Rightarrow s \equiv 2(\bmod 5) . \\
\text { Hence }(s, t)=(2,42),(7,47) .
\end{array}
$$
(ii) If $5 X(t-s)$, then
$$
\begin{array}{c}
25 \text { I }(t+s+1) \Rightarrow t+s+1=25,75 . \\
\text { Hence }(s, t)=(4,20),(8,16),(25,49), \\
(29,45),(33,41) .
\end{array}
$$

In summary, $(2,37,42)$ and $(7,32,47)$ are obtained by $C_{3}$ and $C_{28}$, respectively, and the rest of the people receive at most 2 souvenirs.