Solution:

Note that a chess player could not have more than one draw. Indeed, if $A$ had draws with $B$ and $C$, then the condition for $A$ and $B$ implies that $B$ defeated $C$ and the same condition for $A$ and $C$ implies that $C$ defeated $B$, a contradiction.

Let $A_{1}, A_{2}, \ldots, A_{k}$ be the longest sequence such that each chess player has defeated the next one, i.e. $A_{i}$ has defeated $A_{i+1}$ for $i=1,2, \ldots, k-1$. If $k=2005$, then we have the required sequence. Assume that $k<2005$ and consider a chess player $B$ who is not amongst $A_{1}, A_{2}, \ldots, A_{k}$.

If $B$ has defeated $A_{1}$ then the sequence $B, A_{1}, A_{2}, \ldots, A_{k}$ of length $k+1$ has the above property, which is impossible.

If $A_{1}$ has defeated $B$, then $B$ and $A_{2}$ had not draw because $A_{1}$ has defeated both. If $B$ has defeated $A_{2}$ then the sequence $A_{1}, B, A_{2}, \ldots, A_{k}$ of length $k+1$ has the above property, a contradiction. Therefore $A_{2}$ has defeated $B$. We see analogously that all players $A_{3}, A_{4}, \ldots, A_{k}$ have defeated $B$. Then we obtain again a contradiction by considering the sequence $A_{1}, A_{2}, \ldots, A_{k}, B$ of length $k+1$.

The above argument shows that outside the sequence $A_{1}, A_{2}, \ldots, A_{k}$ there is only one chess player $B$, and $A_{1}$ and $B$ made a draw. Then this is the only draw of $B$. If $A_{2}$ has defeated $B$, then we obtain as above that $B$ has lost from $A_{i}$ for $i=3,4, \ldots, k$ and we have again sequence $A_{1}, A_{2}, \ldots, A_{k}, B$ of length $k+1$. Therefore $A_{2}$ has lost from $B$ and the same holds for $A_{i}, i=3, \ldots, k$.

On the other hand, there is at least one more draw, for example between $A_{i}$ and $A_{j}$. But $A_{i}$ and $A_{j}$ have lost from $B$, which is a contradiction.