Obviosly $ n=2^2*3^2*5*7*11*13$ (because $ n>11*144$) second conditions satisfyed.

Out of curiosity, how do we have 10 consecutive positive divisors if $ 8\not|\ n$?

Peter wrote: Out of curiosity, how do we have 10 consecutive positive divisors if $ 8\not|\ n$? I undestand second condition: exist divisors $ d_{k + 1} > d_k + 10.$ Among divisors $ d_{k}$ and $ d_{k + 1}$ had 10 consequitive numbers. If we have 10 consequitive divisors, then $ n=2^3*3^2*5^2*7*11$.

(i.) $ n$ has exactly 144 distinct positive divisors, (ii.) there are ten consecutive integers among the positive divisors of $ n$. Say the 10 consecutive divisors in (ii.) are $ k,~k + 1,\ldots,~k + 9$. Then, $ 2^3,3^2,5,7$ all divide the $ \mbox{lcm}(k,k + 1,\ldots,k + 9).$ So, in order to satisfy (i.) and (ii.), we require that $ n = 2^{\nu_2}3^{\nu_3}5^{\nu_5}\cdots$ or i.e. $ n = \prod_{p\mid n} p^{\nu_p}$ where $ \nu_p\ge 0$ for all primes $ p \mid n$, and moreover $ \nu_2\ge 3$, $ \nu_3\ge 2$, $ \nu_5\ge 1$, and $ \nu_7 \ge 1$. Let $ N = 2^5*3^2*5*7*11$ and note that $ N$ satisfies (i.) and (ii.). As $ 144 = 2^4 * 3^2$, it follows that $ \nu_p$ can only be $ 1,2,3,5,7,8,11,\ldots$ for each $ p$. Notice that $ 2^{11} * 3^2 * 5 * 7 > N$ implies that $ \nu_2\le 8$ which in turn forces $ \nu_p\le 8$ for each $ p$. This greatly reduces the number of possibilities for the exponents of $ n$. Yet, there is still work to be done. From here one needs to argue that the choice of $ N$ above is least possible. Altogether, we have that $ \nu_2\le 8$ and one can similarly obtain that $ \nu_3\le 5,\nu_5\le 3,\nu_7\le 2,$ and $ \nu_{11}\le 1$. Also, we claim that no prime $ > 11$ can divide $ n$. Keep in mind that $ 2^3*3^2*5*7 \mid n$. Suppose there was such an $ n$ smaller than our choice $ N$ and a prime $ p > 11$ such that $ p\mid n$, then in order to satisfy (ii.): \[ 2^3*3^2*5*7*p^{\nu_p} \le N \mbox{~~i.e~~} p^{\nu_p} \le 44 \] in which we obtain that $ \nu_p = 1$. But in order to satisfy also (i.), we now have the stronger statement that \[ 2^3*3^2*5^2*7*p \le N \mbox{~~i.e~~} p \le 44/5 < 9. \] Thus, the claim follows. Altogether, the only possibilities are $ \nu_2 = 3,5,7,$ or $ 8$, $ \nu_3 = 2,3,$ or $ 5$, $ \nu_5 = 1,2,$ or $ 3$, $ \nu_7 = 1$ or $ 2$, and $ \nu_{11} = 0$ or $ 1$. Simple (yet tedious) analysis gives the above choice $ N$ to be the desired solution.