$\binom{2021}{1010}-1$ For each subset $S$, we consider a path as follows: on the $i$th step, if $i\in S$ then it goes up, otherwise it goes right. Rephrasing, a set $S$ is bally if and only if it stays below the line $y=x+1$. This is necessary because if $y\ge x+2$ at a point where I just went up to $(x,y)$ then before going up I have y at least $\frac{x+y}{2}$, so at least this many elements less than $x+y$ are in $S$. This is sufficient because before going up, $y\le x$ which means there are at most $\frac{x+y}{2}<\frac{x+y+1}{2}$ elements. Extend the path to start at $(-1,0)$ and never goes over the $y=x+1$ line (so it must go to (0,0)) Let $n=1010$. It must end at one of $(n,n), (n+1, n-1), \cdots, (2n,0)$ Claim: I have $\binom{2n+1}{k}-\binom{2n+1}{k-1}$ paths from $(-1,0)$ to $(2n-k,k)$ for all $0\le k\le n$ This solves the problem because the answer telescopes to $\binom{2n+1}{n}-\binom{2n+1}{0}$ (since the empty set, or a horizontal line doesn't count) Shift everything right by 1. Now, there are $\binom{2n+1}{k}$ paths in total, not counting those that go over $y=x$. If a path goes over the line, reflect the part after $(x_0, x_0+1)$, the earliest place it goes above the y=x line over the line $y=x+1$. This creates a bijection from illegal paths to paths to $(k-1, 2n+1-k)$ because after going up, the reflected part contains contains $2n+2-2k$ more ups than lefts, and furthermore we can reflect back based on the earliest instance when the path goes above the x-axis.