1. Find a triplet of integers $(l, m, n)(1<l<m<n)$, such that $\sum_{k=1}^{l} k 、 \sum_{k=l+1}^{m} k 、 \sum_{k=m+1}^{n} k$ form a geometric sequence.
Day 1
1. For $t \in \mathbf{N}_{+}$, let $S_{t}=\sum_{k=1}^{t} k=\frac{t(t+1)}{2}$. Given that $\sum_{k=1}^{l} k=S_{l}, \sum_{k=l+1}^{m} k=S_{m}-S_{l}, \sum_{k=m+1}^{n} k=S_{n}-S_{m}$ form a geometric sequence, then
$$
\begin{array}{l}
S_{l}\left(S_{n}-S_{m}\right)=\left(S_{m}-S_{l}\right)^{2} \\
\Rightarrow S_{l}\left(S_{n}+S_{m}-S_{l}\right)=S_{m}^{2} \\
\Rightarrow S_{l} \mid S_{m}^{2} \\
\Rightarrow 2 l(l+1) \mid m^{2}(m+1)^{2} .
\end{array}
$$

Let $m+1=l(l+1)$, take $l=3$. Then
$$
m=11 \text {. }
$$

Thus, $S_{l}=S_{3}=6, S_{m}=S_{11}=66$.
Substituting into equation (1) gives $S_{n}=666$, i.e.,
$$
\frac{n(n+1)}{2}=666 \Rightarrow n=36 \text {. }
$$

Therefore, $(l, m, n)=(3,11,36)$ is a solution that satisfies the conditions.