Answer: $2310$. Suppose $a$ and $b$ satisfy the problem conditions. For each integer $k$ from $1$ to $10$ denote by $p_k$ the minimal prime divisor of $\gcd(a+k, b+k)$. Since the $\gcd$ of numbers divides their difference, among $10$ successive integers at most $\lfloor \frac{10}{p_k} \rfloor$ can be a multiple of $p_k$. For $p_k = 2$ this number is $5$, for $p_k = 3$ it is $4$, for $p_k = 5$ and $p_k = 7$ it is $2$ and for the others it equals $1$.
If all $p_k$ are odd then the equality $10 = 4 + 2 + 2 + 1 + 1$ implies that there is at least five distinct primes. Then $b - a \ge 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 > 2310$. Consider the case when some $p_i = 2$. Since $a$ and $b$ are coprime, $i$ cannot be even so $2 = p_1 = p_3 = \dots = p_9$. Among five primes $p_2, p_4, \dots, p_{10}$ at most $2$ can be equal to $3$, and for each prime $p \ge 5$ at most one can be equal to $p$. Now the equality $10 = 5 + 2 + 1 + 1 + 1$ implies that $b - a \ge 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$.
On the other hand, if $a = 2311$ and $b = 4621$ then $p_1 = p_3 = \dots = p_9 = 2$, $p_2 = p_8 = 3$, $p_4 = 5$, $p_6 = 7$ and $p_{10} = 11$, whence $|a - b| = 2310$ is possible.