For $x >0$, the two are equal if and only if there exists $n$ such that $n \leq \frac x{20} < \frac x{19} < n+1$. This translates to $20n \leq x < 19n+19$.
Now for a given $n$, we have $(19n+19)-20n = 19-n$. Therefore we have $19-n$ values between $20n$ and $19n+19$ for $n < 19$ which $x$ can take (note : for $n=0$ we actually can't have $x=0$ so we must subtract $1$), and none for $n \geq 19$. Thus, the answer is :
$$
\sum_{n=0}^{18} (19-n) - 1 = 361 - 171 - 1 = 190-1 = 189
$$
In general case of $\lfloor x/n \rfloor = \lfloor x/m \rfloor$ with $n > m$, we get that for $x>0$ equality works if and only if for some $N$ we have $N \leq \frac xn < \frac xm < N+1$. This translates to $nN \leq x < mN + m$.
Therefore for all $N$ such that $N \geq 0$ and $N < \frac{m}{n-m}$, we get $(mN - nN + m)$ values of $x$. So the answer would be (again accounting that $x=0$ will be counted in the $N=0$ case so we must remove it by subtracting $1$):
$$
\sum_{N=0}^{\lfloor \frac{m}{n-m} - 1\rfloor} (mN - nN + m) - 1
$$
in our case, $m=19$, $n= 20$, so $\frac{m}{n-m} - 1 = 18$ ,with $mN - nN + m = 19-n$, so we are back. You can of course evaluate the above sum better ,this I leave to you.
Best Answer
The equation you've given returns the maximum of two numbers; that is,
$$\max\{a, b\} = \frac{a + b + |a - b|}{2}$$
To see this, simply note that if $a \ge b$, we have
$$a + b + |a - b| = a + b + (a - b) = 2a$$
and if $a < b$,
$$a + b + |a - b| = a + b - (a - b) = 2b$$ Then to find the largest of three numbers, simply compare the first two, and then compare the result with the third number.