Problem 806

(Method One)
The 1st blackened cell is 1;
The 2nd blackened cell is $3=1+2$;
The 3rd blackened cell is $6=1+2+3$;
The 4th blackened cell is $10=1+2+3+4$;
$\qquad$
And so on, the $n$-th blackened cell is $1+2+\cdots+n$.
Thus, we can get the blackened cells in sequence as $1 \equiv 1(\bmod 8), 1+2=3 \equiv 3(\bmod 8)$,
$$\begin{array}{l} 1+2+3=6 \equiv 6(\bmod 8), 1+2+3+4=10 \equiv 2(\bmod 8) \\ 1+2+\cdots+5=15 \equiv 7(\bmod 8), \quad 1+2+\cdots+6=21 \equiv 5(\bmod 8) \\ 1+2+\cdots+7=28 \equiv 4(\bmod 8), \quad 1+2+\cdots+8=36 \equiv 4(\bmod 8) \\ 1+2+\cdots+9=45 \equiv 5(\bmod 8), \quad 1+2+\cdots+10=55 \equiv 7(\bmod 8) \\ 1+2+\cdots+11=66 \equiv 2(\bmod 8), \quad 1+2+\cdots+12=78 \equiv 6(\bmod 8) \\ 1+2+\cdots+13=91 \equiv 3(\bmod 8), \quad 1+2+\cdots+14=105 \equiv 1(\bmod 8) \\ 1+2+\cdots+15=120 \equiv 0(\bmod 8) \end{array}$$

Up to this point, the numbers that are $\bmod 8$ with remainders $0 \sim 7$ have all appeared; that is, every column contains at least one black cell;
The number in the last blackened cell is 120.
(Method Two)
The 1st blackened cell is 1;
The 2nd blackened cell is $3=1+2$;
The 3rd blackened cell is $6=1+2+3$;
The 4th blackened cell is $10=1+2+3+4$;
$\qquad$
And so on, the $n$-th blackened cell is $1+2+\cdots+n$.
The blackened cells $\bmod 8$ have remainders in sequence as $1,3,6,2, 7, 5, 4, \ldots$
Up to this point, except for $\bmod 8$ with remainder 0, all other $\bmod 8$ with remainders $1 \sim 7$ have already appeared;
Therefore, we only need to consider $1+2+\cdots+n \equiv 0(\bmod 8)$;
$$\begin{array}{l} 8 \left\lvert\, 1+2+\cdots+n=\frac{1}{2} n(n+1)\right. \\ 16 \mid n(n+1) \end{array}$$

Since $n$ and $n+1$ are coprime;
Thus, $16 \mid n$ or $16 \mid n+1$;
We hope that $n$ is as small as possible;
Then take $n+1=16$;
$$n=15$$

Thus, the number in the last blackened cell is $1+2+\cdots+15=(1+15) \times 15 \div 2=120$.