How many handshakes?

Since one friend shook everyone's hand, that person is effectively not contributing any information and can be removed, while decreasing everyone else's count by 1 (including Alice's). That leaves counts 0 through 5, plus Alice. Now one person shook no one's hand, and can be removed, leaving 1 through 5, plus Alice. That person with "5" shook everyone's hand, and can be removed, etc. Do this once more to get two people, Alice and one other person, and the other person only shook 1 hand (Alice's). That means Alice shook one hand. Undo the three steps earlier (where we subtracted 1 from Alice each step), adding 3 to Alice, and you get 4 for Alice.

Friends represented by letters: A, B, C, D, E, F, G and H. In parentheses are the friends that have shaked hands with that specific friend. Then the total shakes for that specific friend is shown:

• A (BCDEFGH) = 7 shakes.
• B(ACDEFG)=6 shakes.
• C(ABDEF)=5 shakes
• D(ABCE)=4 shakes.
• E(ABCD)=4 shakes.
• F(ABC)=3 shakes.
• G(AB)=2 shakes.
• H(A)=1 shake.

The only number of shakes repeated is 4, so the answer is 4. All the other ones are discarted, as they are unique.

If someone's answer were " $\textbf{0}$ " that means the number " $\textbf{7}$ " will disappear and the condition $\textbf{positive}$ would disappear as well, and if $\textbf{Alice}$ was to answer "0" that means there are at least 2 of her 7 friends will have the same number of handshakes

Now we know for sure the friends hold the numbers from 1 to 7 for handshakes and Alice holds a repeated number to one of them

So, let's name her friends $\color{#D61F06}{\textbf{f1}}, \color{#3D99F6}{\textbf{f2}}, \color{maroon}{\textbf{f3}}, \color{magenta}{\textbf{f4}}, \color{lime}{\textbf{f5}}, \color{#20A900}{\textbf{f6}}, \color{#EC7300}{\textbf{f7}}$

and assume that handshakes $\color{#D61F06}{\textbf{1}}, \color{#3D99F6}{\textbf{2}}, \color{maroon}{\textbf{3}}, \color{magenta}{\textbf{4}}, \color{lime}{\textbf{5}}, \color{#20A900}{\textbf{6}}, \color{#EC7300}{\textbf{7}}$ distributed respectfully

meaning $\color{#D61F06}{\textbf{f1}}$ made $\color{#D61F06}{\textbf{1}}$ hand shake and $\color{#3D99F6}{\textbf{f2}}$ made $\color{#3D99F6}{\textbf{2}}$ handshakes, so on and so forth

If $\textbf{Alice}$ has 7 hand shakes as $\color{#EC7300}{\textbf{f7}}$ that would result in the disappearance of the number " $\textbf{1}$ " as an answer (which means a contradiction to the 7 distinct numbers condition)

because $\textbf{Alice}\space and\space \color{#EC7300}{\textbf{f7}}$ each would have been shook hands with the remaining $\textbf{6}$ friends, meaning the other $\textbf{6}$ friends each would have been shook hands "twice"

The same applies for $\textbf{6}$ and $\textbf{5}$ will result in the disappearance of the numbers " $\textbf{2}$ " and " $\textbf{3}$ "

So the only number remains is $\Large\color{#3D99F6}{\boxed{4}}$

It's like if the 7 friends shook hands together with respect to the condition

And the result was

$\color{#D61F06}{\textbf{f1=1}}, \color{#3D99F6}{\textbf{f2=2}}, \color{maroon}{\textbf{f3=3}}, \color{magenta}{\textbf{f4=3}}, \color{lime}{\textbf{f5=4}}, \color{#20A900}{\textbf{f6=5}}, \color{#EC7300}{\textbf{f7=6}}$

And then came $\textbf{Alice}$ and shook hands with $\color{magenta}{\textbf{f4}}, \color{lime}{\textbf{f5}}, \color{#20A900}{\textbf{f6}}, \color{#EC7300}{\textbf{f7}}$ raising them up $\textbf{+1}$ handshake

Since handshakes involve two persons, the total number of handshakes by everyone involved will be an even number - zero is even, and every handshake increases the count by one for two persons. The sum of 1 + 2 + 3 + 4 + 5 + 6 + 7 is 28, so we know that Alice must have shaken an even number of hands. (We already knew that she didn't shake them all (7), since all of her other friends shook hands with all distinct, positive , numbers - including seven, since she has seven friends here - which means that someone only shook one person's hand...and it wasn't hers; sounds kind-of rude for a "friend," but at least that person came to the party.)

So, Alice shook either two, four or six hands. It couldn't have been six, for similar reasons to why it couldn't have been seven: since two other people shook seven and six hands, there can only be one person who shook one hand, and one person who shook only two hands. Hey, by that argument, that means that she couldn't have only shaken two hands, either.

Thus, Alice must have also shaken four hands .