Harmonic Triangle
Next two rows:
$$\frac{1}{7}, \frac{1}{42}, \frac{1}{105}, \frac{1}{140},
\frac{1}{105}, \frac{1}{42}, \frac{1}{7}$$
$$\frac{1}{8}, \frac{1}{56}, \frac{1}{168}, \frac{1}{280},
\frac{1}{280}, \frac{1}{168}, \frac{1}{56}, \frac{1}{8}$$
Pattern: the second number in the $n$th row is
$\frac{1}{n(n-1)}$.
We can always fill the
second diagonal, because
$$\frac{1}{n}+\frac{1}{n(n-1)}=\frac{[(n-1)+1]}{[n(n-1)}=\frac{1}{(n-1)}$$
And this is clearly a unit fraction.