Consider the number of teams $n$ who have previously played an odd number of games. $n$ must always be even, since each game adds 1 to the previous-game-count of two teams (aka handshake theorem). Each odd game doesn't affect $n$; each non-odd game increases or decreases $n$ by $2$. Furthermore at the beginning and the end of the tournament, we have $n = 0$. Thus, by taking $n \bmod{4}$ we find that the number of non-odd games must have been even. The total number of games is $\binom{15}{2} = 105$, which is odd; thus, the number of odd games must be odd, and cannot be zero.

Yes. We will construct one such tournament inductively, starting with $3$ teams and adding $4$ teams each time. With $3$ teams it's easy to check that absolutely any valid tournament has exactly one odd game. Now, given a tournament for $t$ teams ($t$ must be odd), we'll add four teams $A, B, C, D$ without adding any odd games: First, play the original $t$-team tournament. Every team has still played an even number of games. Now play $AC, BD, AD, BC, AB$ in that order, so that $A$ and $B$ have both played an odd number of games, and $C$ and $D$ have both played an even number of games. Label the other teams $T_1$ to $T_t$, and play $CT_{2k-1}, AT_{2k-1}, DT_{2k-1}, BT_{2k-1}, AT_{2k}, CT_{2k}, BT_{2k}, CT_{2k}$ for $k = 1, 2, \ldots, \frac{t-1}{2}$ and $CT_t, AT_t, DT_t, BT_t$ Finally play $CD$.