Among the list $10,11,12,13,14,15$, the integers 11 and 13 are prime. Also, $10=2 \times 5$ and $12=2 \times 2 \times 3$ and $14=2 \times 7$ and $15=3 \times 5$. For an integer $N$ to be divisible by each of these six integers, $N$ must include at least two factors of 2 and one factor each of $3,5,7,11,13$. Note that $2^{2} \times 3 \times 5 \times 7 \times 11 \times 13=60060$. (This is the least common multiple of $10,11,12,13,14,15$.) To find the smallest six-digit positive integer that is divisible by each of $10,11,12,13,14,15$, we can find the smallest six-digit positive integer that is a multiple of 60060. Note that $1 \times 60060=60060$ and that $2 \times 60060=120120$. Therefore, the smallest six-digit positive integer that is divisible by each of $10,11,12,13,14,15$ is 120120. The tens digit of this number is 2.