What is the expected number of coin tosses it would take to get N many heads OR N many tails?

The probability that the game ends at round $i$ is:$$\binom{i-1}{N-1}2^{1-i}$$If e.g. the game ends with a tail at $i$-th round then among the $i-1$ results in the former rounds must be $N-1$ tails. There are $\binom{i-1}{N-1}$ configurations for that and each of them has probability $2^{1-i}$ to occur. Further we must multiply with the probability that a tail will appear at $i$-th trial which is $\frac12$. Same outcome if the game ends with a head at $i$-th round so a multiplication with $2$ is also needed.

So to be found is:

$$\sum_{i=N}^{2N-1}i\binom{i-1}{N-1}2^{1-i}=2N\sum_{i=N}^{2N-1}\binom{i}{N}2^{-i}$$