Consecutive!

If $n$ is the smallest number, then one each of $n,n+1,n+2,n+3,n+4,n+5,n+6,n+7$ must be divisible by $2,3,5,7,11,13,17,19$ . Thus, in particular, two of these numbers must be divisible by $17$ and $19$ . Solving the appropriate Chinese Remainder Theorem equations, we see that $n +a \equiv 0 \pmod{17} \;,\; n + b \equiv 0 \pmod{19} \hspace{0.5cm} \Leftrightarrow \hspace{0.5cm} n \; \equiv \; 152a - 153b \pmod{17\times19=323}$ and so it is only possible for two of these numbers to be divisible by $17$ and $19$ if $n$ is congruent to one of the numbers $\begin{array}{cccccccc} 12 & 13 & 14 & 15 & 16 & 17 & 31 & 32 \\ 33 & 34 & 50 & 51 & 95 & 112 & 113 & 114 \\ 129 & 130 & 131 & 132 & 133 & 146 & 147 & 148 \\ 149 & 150 & 151 & 152 & 164 & 165 & 166 & 167 \\ 168 & 169 & 170 & 183 & 184 & 185 & 186 & 187 \\ 202 & 203 & 204 & 221 & 265 & 266 & 282 & 283 \\ 284 & 285 & 299 & 300 & 301 & 302 & 303 & 304 \end{array}$ modulo $323$ . It is easy to set up an Excel spreadsheet to test these options, and the smallest value of $n$ for which we get the desired divisibility for all eight numbers is $n = 338$ (the second instance of $n \equiv 15 \pmod{323}$ ). This makes the answer $\boxed{2732}$ .

The 8 numbers are from 338 to 345.

$344=2\times172$ $339=3\times113$ $345=5\times69$ $343=7\times49$ $341=11\times31$ $338=13\times26$ $340=17\times20$ $342=19\times18$