Excluding the trivial edgeless case of \(\displaystyle K_1\) we can proceed as follows:
If n is odd
Then \(\displaystyle K_n\) has all even degrees, and so by a theorem of Euler's we have that there exist an Eulerian circuit, which by technicality admits an Eulerian path (a circuit is a kind of path).
If n is even
Well clearly \(\displaystyle K_2\) contains an Eulerian path but not an Euler circuit. However for all even n>2 we know that they have more than 2 vertices of odd degree (for \(\displaystyle K_n\) we have all vertices with degree n-1). For there to be an Eulerian path, we can have at MAX two vertices of odd degree, and for even n>2, this condition fails. And so \(\displaystyle K_2\) is the only graph the has an Eulerian path but no circuit.