【Answer】 120

【Solution】If $m \equiv 1(\bmod 2)$, then the remainder of $m$ divided by $4, 6, 8, 10$ cannot be 0, and the sum of these remainders exceeds 4, so $m \equiv 0(\bmod 2)$. Since the sum of the remainders is 4, for an even number, its remainder when divided by 8 can only be $0, 2$ or 4 (if it is 6, it exceeds 4).
(1) If the remainder of this number when divided by 8 is 4, then it must be a common multiple of $3,5,7,9,10$. Since $[3,5,7,9,10]=630$, to satisfy the remainder of 4 when divided by 8, this number is at least $630 \times 2=1260$;
(2) If the remainder of this number when divided by 8 is 2, then its remainder when divided by 4 is also 2 (all 4 remainders are used up), so it must still be a common multiple of $3,5,7,9,10$, and this number is at least $630 \times 3=1890$;
(3) If the remainder of this number when divided by 8 is 0, then $m \equiv 0(\bmod 2), m \equiv 0(\bmod 4), m \equiv 0(\bmod 8)$. We need to further analyze. If the remainder of $m$ when divided by 3 is not 0, then its remainder when divided by 6,9 will also not be 0. Since $m$ is even, the remainder of $m$ when divided by 6 is at least 2. To make the sum of the remainders 4, it can only be $\left\{\begin{array}{l}m \equiv 1(\bmod 3) \\ m \equiv 2(\bmod 6) \\ m \equiv 1(\bmod 9)\end{array}\right.$, but $m \equiv 2(\bmod 6) \Rightarrow m \equiv 2(\bmod 3)$, which is a contradiction, so this number must be a multiple of 3. Since this is an even number and it is also a multiple of 3, it must be a multiple of 6, so $\left\{\begin{array}{l}m \equiv 0(\bmod 3) \\ m \equiv 0(\bmod 6)\end{array}\right.$ . Up to this point, we have deduced that: $m \equiv 0(\bmod 1), m \equiv 0(\bmod 2), m \equiv 0(\bmod 3)$, $m \equiv 0(\bmod 4), m \equiv ?(\bmod 5), m \equiv 0(\bmod 6), m \equiv ?(\bmod 7), m \equiv 0(\bmod 8)$, $m \equiv ?(\bmod 9), m \equiv ?(\bmod 10)$ (where $?$ represents the undetermined remainders). Next, we discuss the remainder of $m$ when divided by 9:
(3.1) If $m \equiv 3(\bmod 9)$, only one remainder is left. Considering $\left\{\begin{array}{l}m \equiv 1(\bmod 5) \Rightarrow m \equiv 1 \text{ or } 6(\bmod 10) \\ m \equiv 1(\bmod 10) \Rightarrow m \equiv 1(\bmod 5)\end{array}\right.$, so the remaining remainder should be given to 7, that is, $m \equiv 0(\bmod 5)$, $m \equiv 1(\bmod 7)$, $m \equiv 3(\bmod 9)$, $m \equiv 0(\bmod 10)$, at this time $m$ is at least 120;
(3.2) If $m \equiv 0(\bmod 9)$, 4 remainders are left. Since $\left\{\begin{array}{l}m \equiv 0(\bmod 8) \\ m \equiv 0(\bmod 9)\end{array}\right.$, at this time $m$ is already a multiple of $[8,9]=72$. Clearly, 72 does not meet our requirements, and $72 \times 2$ already exceeds 120;
In summary, $m$ is at least 120.
11. For any positive integers $m, n$, define $r(m, n)$ as the remainder of $m \div n$ (for example, $r(8,3)$ represents the remainder of $8 \div 3$, so $r(8,3)=2$). The smallest positive integer solution that satisfies the equation $r(m, 1)+r(m, 2)+r(m, 3)+\cdots+r(m, 10)=4$ is $\qquad$.