13. Define $n!=1 \times 2 \times \ldots \times n$, for example, $5!=1 \times 2 \times 3 \times 4 \times 5$. If $\frac{n!\times(n+1)!}{2}$ (where $\mathbf{n}$ is a positive integer, and $1 \leq n \leq 100$) is a perfect square, for example, when $n=7$,
$\frac{n!\times(n+1)!}{2}=\frac{7 \times (7+1)!}{2}=\frac{7 \times 8!}{2}=\frac{7 \times (7 \times 8)}{2}=(7!)^{2} \times 4=(7!)^{2} \times 2^{2}$ is a perfect square, then the sum of all $\mathbf{n}$ that satisfy the condition is
【Answer】273
Analysis:
$\frac{\mathrm{n}!(\mathrm{n}+1)!}{2}=(\mathrm{n}!)^{2} \times \frac{\mathrm{n}+1}{2}$, let $\frac{\mathrm{n}+1}{2}$ be a perfect square, $\frac{\mathrm{n}+1}{2}=\mathrm{k}^{2}, \mathrm{n}=2 \mathrm{k}^{2}-1$
(1) $\mathrm{k}=1, \mathrm{n}=1$
(2) $k=2, n=7$
(3) $\mathrm{k}=3, \mathrm{n}=17$
(4) $\mathrm{k}=4, \mathrm{n}=31$
(5) $\mathrm{k}=5, \mathrm{n}=49$
(6) $\mathrm{k}=6, \mathrm{n}=71$
(7) $\mathrm{k}=7, \mathrm{n}=97$
The sum of all $\mathrm{n}$ that satisfy the condition is $1+7+17+31+49+71+97=273$