Convert 2N4_seven to base 10.

To convert a number from base $b$ to base 10, we use the formula $\sum_{i=0}^{k} d_i \times b^i$, where $d_i$ are the digits and $k$ is the highest power.

$$2N4_{\text{seven}} = 2 \times 7^2 + N \times 7^1 + 4 \times 7^0$$ $$= 2 \times 49 + N \times 7 + 4 \times 1$$ $$= 98 + 7N + 4$$ $$= 102 + 7N$$

$$15N_{\text{nine}} = 1 \times 9^2 + 5 \times 9^1 + N \times 9^0$$ $$= 1 \times 81 + 5 \times 9 + N \times 1$$ $$= 81 + 45 + N$$ $$= 126 + N$$

Since $2N4_{\text{seven}} = 15N_{\text{nine}}$, we set their base 10 equivalents equal:

$$102 + 7N = 126 + N$$

Subtract N from both sides:

$$102 + 6N = 126$$

Subtract 102 from both sides:

$$6N = 126 - 102$$ $$6N = 24$$

Divide by 6:

$$N = \frac{24}{6}$$ $$N = 4$$

The value of N must be less than the base in which it appears. In base 7, N must be less than 7. In base 9, N must be less than 9. Since $N=4$, it satisfies both conditions.

The value of N is $\textbf{4}$.