Egyptian heritage

The LCM of 2,3,4,5,6 is 60. Thus we want to find integers $N$ such that $N \equiv 0 \pmod{7}$ while $N \equiv 1 \pmod{60}$ . Thus $N = 7a = 60b + 1$ , so that $\begin{aligned} 7a - 60b & = \; 1 \; = \; 2\times 60 - 17 \times 7 \\ 7(a + 17) & = \; 60(b + 2) \end{aligned}$ so that $a + 17 = 60c$ and $b+2 = 7c$ for some integer $c$ , and hence we deduce that $N \; = \; 420c - 119 \hspace{2cm} c \in \mathbb{Z}$ The sum of the 10 smallest positive such integers is $\sum_{c=1}^{10}(420c - 119) \; = \; \boxed{21910}$

Using the same deduction as Mark Henning re 420, but not realizing how to get the 301 number, I just searched for it. $\frac{1}{2} (9\ 10) 420+10\ 301=21910$ . That is, ten copies of 301 plus $\frac{9(9+1)}{2}$ copies of 420. The fraction is the $9^{th}$ triangular number.