Each handshake involves 2 people.
If the ith person shakes hands $n_{i}$ times, then the sum of the $n_{i}$ be even.
If there were an odd number of odd $n_{i}$ then the sum would be odd.
or proof by parity argument as shown :
Let E be the set of people who have shaken hands an even number of times.
Let O be the set of people who have shaken hands an odd number of times.
Let n be the number of times any hand has shaken another.
Let n_o be the number of times a hand from the set O has shaken another.
Let n_e be the number of times a hand from the set E has shaken another.
Note: n = n_o + n_e
Each handshake increases n by two, hence n is even.
n_e is a sum of even numbers (by definition) and hence is also even.
Since both n and n_e are both even, the equation above tells us that n_o is even as well.
But n_o is a sum of odd numbers (by definition). The only way for n_o to be even then, is if the size of O is even.
