An olympiad question on numbers 
Consider the sum of $92 + 105$, so the expression is $197$, however if you turn it upside down, you get $501+ 26$ or $527$ :
\begin{align}
105+92 \to 26+501
\end{align}
Now I'm looking for an expression that when you turn it upside down the sum will increase by exactly 2020 units !!!

But for solving this puzzle we have the following rules :
1- We are just allowed to use digits and the signs $+ , - $
2- None of the $2$ numbers (before and after the spin) can start with $0$
3-The value of the phrase must be positive.

Here is my try :
If we pay attention to the example we'll get some information: the difference between two numbers $92$ and $26$ is $66$ and the difference between $105$ and $501$ is $396$ and also the difference between $197$ and $527$ is $330$ now look at this : $ 396 - 66 = 330$; I guess there must be a relation between this and our answer!

Can anyone help ?
 A: To begin, I assume that you are allowed many numbers and many $+$'s and/or $-$'s in your expression.  If you only allow your expression to involve two numbers and a single $+$ or $-$ then more thought is required.
Let the flip-function be labeled as $F$ which takes in an unevaluated expression and outputs the flipped expression which can then be evaluated.
First, let us notice that if the expression passed in to $F$ contains only addition, that the result has the same remainder modulo $3$ before and after.  This is because the digits themselves who get flipped stay the same remainder class modulo $3$ before and after and rearranging the digits in a summation or in a number itself does not change the remainder modulo $3$.  As such, an expression involving only addition minus the flip of the expression will always result in an expression that evaluates to a multiple of $3$ and could never result in $2020$.
It follows then that for such an expression to exist who when subtracting the flip of the expression evaluates to $2020$ there must be subtraction involved.
Well, $92-15=77$ while $F(92-15)=51-26=25$.  We have $(92-15)-F(92-15)=77-25=52$ which has remainder $1$ when considered modulo $3$, the same equivalence class as $2020$ is.
Now, note that:
$9+9+9+\dots + 9= 9k$
$F(9+9+9+\dots+9)=6+6+6+\dots+6=6k$
$9+9+\dots+9-F(9+9+\dots+9)=9k-6k=3k$
So, we could have the following expression:
$$\underbrace{9+9+9+\dots+9}_{k~\text{copies}} + 92 - 15$$
The flip of the expression:
$$51-26+\underbrace{6+6+6+\dots+6}_{k~\text{copies}}$$
The difference of these:
$52 + 3k$ which by setting $k=656$ is precisely equal to $2020$

With regards to only allowing our initial expression to include two numbers and a single operation, we can limit our search noting what we did earlier that it must be a minus and could not have been a plus.  Writing a computer script can yield some results:
For instance $$6969-2296$$
as well as $$6969-2626$$
Indeed, $6969-2296 = 4673$ and $F(6969-2296)=9622-6969=2653$ which is precisely $2020$ off from the earlier evaluation.
Admittedly, I don't see a clean way to have arrived at these numbers by hand without a great deal of trial and error, but knowing the numbers must have been different equivalence classes modulo $3$ and a minus sign must be used helps narrow it down considerably.

 The admittedly very dirty script: flip = function(exp){ exp = exp.replaceAll("6","n").replaceAll("9","x"); exp = exp.replaceAll("n","9").replaceAll("x","6"); ret = ''; for(i=0;i<exp.length; i++){ret+=exp[exp.length-i-1]} return ret} for(a=5000;a<7000;a++){ for(b=1;b<a; b++){ exp = '' + a + '-' + b; if(exp.contains("3")||exp.contains("4")||exp.contains("7")){continue} else{if(eval(exp +'-('+flip(exp)+')')==2020){console.log(exp)}}}} which searched for solutions with $a$ only in the range $5000$ to $7000$ and ignored limitations on leading zeroes.

