An extremely tedious problem(which costs me and some other IMO 2020 team members $2$ hours during the test). The answer is $2$. We will first describe a "winning strategy" with $2$ operations. Denote Miss Adams and Mr. Bean by $A$ and $B$ respectively. A will first put $$(2,3,...,15,38)\hspace{3pt}\text{and} \hspace{3pt}(37,39,40,41)$$on both sides, afterwards, he will put $$(1,2,37)\hspace{3pt}\text{and}(41)$$on both sides. In $B's$ view, he will see the following: $$a_2+a_3+...+a_{15}+a_{16}=a_{17}+a_{18}+a_{19}+a_{20}$$and $$a_1+a_{2}+a_{17}<a_{20}$$CLAIM. $a_{17},a_{18},a_{19},a_{20}$ are all in the set $\{37,38,39,40,41\}$ Proof. Just a sketch since it is too annoying. Let $C=\{37,38,39,40,41\}$ and $D=\{a_{17},a_{18},a_{19},a_{20}\}$. We can reject the case $|C\cap D|=2$ and $|C\cap D|=3$ based on the value of $a_1$(noticing that $a_1$ must be odd since the total weight of the coins is odd. Each case is just a simple calculation. For example, if $a_1=41$, then $$\frac{315-41}{2}=R.H.S.\leq 15+38+39+40$$which is impossible. $\blacksquare$ Now the only solution to $a_1+a_2+a_{17}<a_{20}$ is $(1,2,37,41)$. Moreover from $a_{1}$ is odd $B$ will know that $a_1=1$. $\blacksquare$ $\noindent\rule{15.5cm}{0.4pt}$ Now we will show that $1$ operation is not sufficient, which is just more annoying(so I will just provide a sketch). CASE I: $1$ is weighted 1. If there are at least two coins on the same side as $1$, then we can't distinguish that coin and $1$, contradiction. 2. Hence $1$ is the only coin in that side, swapping $1$ and $2$ the expression still holds, so we can't distinguish $1$ and $2$, contradiction. CASE II: $1$ is not weighted 1. Firstly if it is an inequality then swapping $1$ with any coin on the weaker side the inequality still holds, so we can't distinguish $1$ and that coin, contradiction. 2a. So it must be an equality. If there are more than one unweighted coin then we can't distinguish between $1$ and that coin, contradiction. 2b. So exactly $19$ coins are weighted. Now let the set of coins in L.H.S. be $X$ and the set of coins in R.H.S. be $Y$. Suppose $3\in X$, we have that if $n\in X$ then $n\in X+1$, otherwise we can swap $1,3$ and swap $n,n+1$, so we can't distinguish between $1$,$3$, contradiction. Similarly we have that $5\in X$, so if $n\in X$ then $n+2\in X$. Now $3,5,6,...,15\in X$, by counting the total weight we have that the other elements in $X$ is $2,4,40$, but this implies $41$ is not in $X$, contradiction. This completes the proof of every case and we are done.