Unfortunately, an exact answer will depend on the specific sequence of $N$ symbols. To see why, take a simple example in which you only have two possible symbols, A and B, with $N = 2$, and only one monkey. Compare the expected times until the monkey types the sequence AA vs. the sequence BA. Unless the monkey types AA at the very beginning (with probability $\frac{1}{4}$), the first time AA appears the BA sequence must have already occurred. So the latter sequence will have the shorter expected time.
There are lots of these sorts of counterintuitive results about the likelihood of seeing a certain sequence before another certain sequence and about the expected time until a sequence is first seen. For more information, see MathWorld's entry on Coin Tossing, or, for even more information, the article "Penney Ante" that appeared in the UMAP Journal.
Now, if $N$ is large and the symbols are assumed to be evenly distributed in the target sequence, maybe you can avoid the problems with the two-symbol and unevenly distributed example I gave here. Or, if you force each monkey to stop after typing exactly $N$ symbols, check to see if they have typed the target sequence, and then have them start over again from scratch if they have not, then you can definitely avoid these problems. (For more on the latter, see this article "Infinite Monkey Business".)