Mario's Staircase, Again

We can use the formula for the sum of terms of an arithmetic progression .

Setting $a_1=1$ and $d=1$ , we have

$S=\dfrac{n}{2}\left[2a_1+(n-1)d\right]$

$50=\dfrac{n}{2}\left[2(1)+(n-1)(1)\right]$

$100=n(2+n-1)$

$100=n(1+n)$

$100=n+n^2$

$n^2+n-100=0$

Using the quadratic formula to solve for $n$ we get,

$n=9.512...$

It means that the maximum number of complete steps that can be built is $\boxed{9}$ .

If there are $k$ steps, then the number of blocks we need is: $1+2+3+\dots+k=\dfrac{k*(k+1)}{2}$

For $k=9$ , $\dfrac{9*10}{2}=45$ , but for $k=10$ , $\dfrac{10*11}{2}=55$ , so the maximum possible number of steps is $\boxed{9}$ .