A problem by Zi Song Yeoh

$2013 = 3 \times 11 \times 61 = 33 \times 61$

So, if we can attain at least $94$ then we are done on the next minute . Confused with at least ? Just note that from $n + 1$ we can easily attain $n$ . Thus, we have to rule out positive integers from which we can't go greater than $93$ .

Consider $n \leq 4$ . For this case we can't get integers greater than $4$ . And for every $n \geq 5$ we can hop and jump to reach integers greater than $93$ and then $94$ by the method . It concludes that we can attain $2013$ if $5 \leq n \leq 1000$ .So, the answer is $1000 - 4 = \boxed{996}$

Note that $2013=3\cdot 11\cdot 61=33\cdot 61$ , so $33+61=94$ can make 2013.

Also, if $n>4$ , then $2(n-2)=2n-4>n$ is achievable, so $n$ can be increased arbitrarily.

In addition, $1(n-1)=n-1<n$ is achievable, so $n$ can be decreased arbitrarily.

Then, any $4<n\le 1000$ works. Since 1,2,3,4 cannot be increased past themselves, the answer is $\boxed{996}$ .