We first compute the entries of the following matrix

We observe that since $k!$ is divided by $100$ for $k \geq 10$, the sum $1!+2!+\cdots+k!$ leaves a remainder $13$ when divided by $100$ for $k \geq 9$. If equality holds
$$1! + 2! + \cdots + k! = 1 + 2 + \cdots + n = \frac{n(n+1)}{2}$$
for some $k \geq 10$, then there is some natural number $m$ such that
$$n(n + 1) = 100m + 26$$
or
$$n^2 + n - 100m - 26 = 0.$$
The discriminant is $\Delta = 400m + 105$ and it should be a perfect square. On the other hand we cannot have
$$x^2 \equiv 5 \pmod{100},$$
as otherwise $25$ would divide $5$. Therefore the given relation cannot hold for $k \ge 9$, as well as, for $k = 7$. Since
$$9 = 3^2, \quad 153 = 3^2 \cdot 17, \quad 873 = 3^2 \cdot 97 \text{ and } 46233 = 3^2 \cdot 11 \cdot 467,$$
we observe that $2 \cdot 153 = 18 \cdot 17$, a product of two consecutive integers $k > 2$. Therefore the only solution is $(k, n) = (5, 17)$.