Solution:

The winning strategy can be determined by working backwards. Whoever leaves only one match on the table loses. Whoever leaves two wins, because they force the other player to leave only one. Whoever leaves 3 or 4 loses, because on the following move the other player can leave two. Whoever leaves 5, on the other hand, wins, because they force the other player to leave 3 or 4, which we have seen to be losing moves. Similarly one sees that leaving $6,7,8,9$ or $10$ is a losing move (it allows the other player to leave $5$), while leaving $11$ is winning.

This reasoning can be repeated indefinitely: if leaving $k$ matches is a winning move, then leaving a number between $k+1$ and $2k$ (inclusive) is losing, while leaving $2k+1$ is winning. In conclusion the "winning numbers" can easily be determined starting from two and exploiting recursively the fact just observed that if $k$ is winning, then $2k+1$ is also winning. The "winning numbers" less than $1999$ are therefore: $2, 5, 11, 23, 47, 95, 191, 383, 767, 1535$. Barbara can therefore win by following this strategy: on the first move she leaves $1535$ matches on the table, the second time it is her turn she leaves $767$, the third time $383$, and so on. Proceeding in this way, the tenth time it is her turn she will leave $2$ on the table, and on the following move Alberto will lose.