About $n$

$n + (n-1) + ... + 2 + 1 + 2 + ... + (n-1) + n = 461$

$n + (n-1) + ... + 2 + 2 + ... + (n-1) + n = 460$

$2(n + (n-1) + ... + 2) = 460$

$n + (n-1) + ... +2 = 230$

$\frac{n(n+1)}{2} - 1 = 230$

$\frac{n(n+1)}{2} = 231$

$n^2 + n = 462$

$n^2 + n - 462 = 0$

$(n + 22)(n - 21) = 0$

$n = 21$

So, $n - 1 = \boxed{20}$

$n+(n-1)+\ldots + 2 + 1 + 2 + \ldots + (n-1) + n = 461\\ 1+2 + \ldots + (n-1) + n + 2 + \ldots + (n-1) + n = 461\\ 1 + 2 + \ldots + (n-1) + n + 1 + 2 + \ldots + (n-1) + n = 461 + 1\\ \displaystyle 2 \sum_{i=1}^n i = 462\\ 2\dfrac{(n)(n+1)}{2} = 462\\ n^2 + n - 462 = 0\\ (n+22)(n-21)=0\\ n=21,-22$

From the expression, we know that the sequence is $1+ 2 + \ldots + n$ , so we know that $n>0$

Therefore, $n=21$ and

$n-1 = 21 - 1 = \boxed{20}$

Take the square root of 461, floor it, and subtract one.

EDIT: I have found out why it works.

Take a square. Let the sides be of length x.

$Area = x^2$

We can find the area also as a summation:

$1 + 3 + 5 + 7 + ... + x = (\lfloor\frac{x}{2}\rfloor + 1)^2$

which we can split into:

$1 + (1 + 2) + (2 + 3) + (3 + 4) + ... + \lfloor\frac{x}{2}\rfloor + (\lfloor\frac{x}{2}\rfloor +1) = (\lfloor\frac{x}{2}\rfloor + 1)^2$

Manipulating the equation,

$(\lfloor\frac{x}{2}\rfloor +1) + \lfloor\frac{x}{2}\rfloor + \ldots + 2 + 1 + 2 + \ldots + \lfloor\frac{x}{2}\rfloor + 1 = (\lfloor\frac{x}{2}\rfloor + 1)^2$

$(\lfloor\frac{x}{2}\rfloor +1) + \lfloor\frac{x}{2}\rfloor + \ldots + 2 + 1 + 2 + \ldots + \lfloor\frac{x}{2}\rfloor = (\lfloor\frac{x}{2}\rfloor + 1)^2 - 1$

Letting $\lfloor\frac{x}{2}\rfloor = m$ , $m + 1 = n$ :

$(m + 1) + m + \ldots + 2 + 1 + 2 + \ldots + m + (m + 1) = (m + 1)^2 + (m + 1) -1$

$461 = (m + 1)^2 + m$

Now we see that if we square root this equation, we get an ugly mess of irrationality behind (m + 1) due to there being an extra '+ m' behind the equation. There we floor the answer to get rid of it.

Now we got $m + 1 = 21$ , or $n = 21$ .

Therefore $\boxed{n - 1 = 20}$ .

$The\quad equation\quad can\quad be\quad written\quad as,\quad \\ 1+2\quad \times (2+3+4+5+.....+n)\quad =\quad 461\\ 2\times (2+3+4+5+....+n)\quad =\quad 460\\ 2\times (1+2+3+4+5+....+n-1)\quad =\quad 460\\ 2\quad \times \quad (\frac { n(n+1) }{ 2 } -1)=460\\ 2\quad \times \quad (\frac { n(n+1)-2 }{ 2 } )=460\\ n(n+1)-2=460\\ n(n+1)\quad =\quad 462\\ n(n+1)\quad =\quad 21\quad \times \quad 22\\ On\quad comparing,\\ n=21\\ \\ n-1=21-1=20$