I claim that for every $m$, there is a natural number $N$ such that it's representation in all the bases $b$, $2\le b\le 1389$ has at least $m$ non-zero terms. suppose the contrary, meaning that for some $m+1$, every natural number $N,1\le N\le 2^n-1$ (for some large $n$) coflicts our assumption for some base $b$, that is it has at most $m$ non-zero digits in base $b$. now we count them in another way. let $A_i$ be the number of numbers in base $i$ $(2\le i\le 1389)$ with at most $n$ digits that have at most $m$ non-zero digits. obviously $A_i=\dbinom{n}{m}(i-1)^m+...+\dbinom{n}{2}(i-1)^2+\dbinom{n}{1}(i-1)$. now since each natural number $N$ less than $2^n-1$ conflicts our assumption for at least one $i$, $2\le i\le 1389$, we have $\sum_{i=2}^{1389}A_i\ge 2^n-1$. but this is a contradiction, since the left hand side is a polynomial of degree $m$ in terms of $n$, but the right hand side is a power function in terms of $n$, and it's growth is more than the left hand side for sufficient large $n$.

My solution: Let $2*3*...*1989$ be $M$ Denote $T_{b}(m)$ be the number of digits of $m$ in base $b$, for instance: $m=\overline{x_k...x_1}_{b}$ then $T_{b}(m)=k$ Let us consider the number $W$ as follow: $W=M^{k_s}+M^{k_{s-1}}+...+M^{k_1}$ with $s>m$ such that $k_1<k_2<...<k_s$, of course, we have to find sufficient conditions for $k_1,k_2,...,k_s$ we choose $k_1,...,k_s$ inductively First, we choose $k_1$ arbitrarily, then we choose $k_2$ as follow: $k_2>max(T_{b}(\dfrac{M^{k_1}}{b^{k_1}}))+k_1$ for all $b=2,3,...,1989$ then we can easily see that the digits of base $b$ of $M^{k_2}$ are all independent of $M^{k_1}$ that is $M^{k_2}=\overline{x_i...x_100...0}_{b}$ with at least $k_2$ digits zero at the end, then $M^{k_1}+M^{k_2}=\overline{x_i...x_10...0y_j...y_1}_{b}$ with $\overline{y_j...y_1}=M^{k_1}$. Since $x_i,y_j$ are nonzero digits, thus $M_{k_1}+M_{k_2}$ has at least 2 nonzero digits in base $b$, for all $b=2,3,...,1989$, inductively, we will have $W$ has at lease $s$ nonzero digits in base $b$ for all $b=2,3....,1989$ then the only thing we have to do is to choose $s>m$ thus $S_{b}{W}\geq s>m$ QED

Every variable in the solution are positive integers otherwise stated. Let $S(n,b,m) = \{ p | n \geq p \geq 1 \text{ and } s_b(p) \leq m \}$, and $T(n,b,m) = \{ p | n \geq p \geq 1 \text{ and } s_b(p) > m \}$. Note that $|S(n,b,m)|+|T(n,b,m)| = n$ for any $b,m$. Lemma - 1 : For any fixed $b > 1$, $m \geq 0$, we have $\lim_{n \rightarrow \infty} \frac{|S(n,b,m)|}{n} = 0$. ProofLet $f_b(i,j)$ denote the numbers which takes atmost $i$ digits to write in base $b$ and has digit sum exactly equal to $j$. Let $g(i,j)$ denote the number of solutions to the equation $$ x_1 + x_2 + \cdots + x_i = j$$with $x_i \in \mathbb{N}_{0}$. Now note for any $i,j$, we have $f_b(i,j) \leq g(i,j)$ (assume $\max(x_1, \cdots, x_i) \leq b-1$, then interpret $x_1x_2 \cdots x_i$ in base $b$). Also note that $g(i,j) = \binom{i+j-1}{j} = \frac{(i+j-1)(i+j-2)\cdots(i-1)}{j!} < (i+j-1)^j$. Now, WLOG assuming $n > b^{{m+100000}^{m+10000}}$ and letting $i = \lfloor \log_b n \rfloor + 2$ $$ 0 \leq \frac{|S(n,b,m)|}{n} \leq \frac{g(i,0) + g(i,1) + \cdots + g(i, b-1)}{n} < \frac{\sum_{0 \leq j \leq b-1} (i+j-1)^j}{n}$$$$< (i+m-1)^m \frac{m+1}{n} < \frac{(2i)^m}{n} (m+1) \leq \frac{(100 \log_b n)^m (m+1)}{n}$$. Now since $\lim_{n \rightarrow \infty}\frac{(\log_b n)^m}{n} = 0$, by squeeze theorem and noting that $100^m(m+1)$ are constant, we see $\lim_{n \rightarrow \infty} \frac{|S(n,b,m)|}{n} = 0$. Lemma - 2 Let $P_1, P_2, \cdots, P_k \subset \{1, \cdots, n \}$. Then $|P_1 \cap P_2 \cap P_3 \cdots P_k| \geq \max(0, |P_1|+|P_2|+\cdots+|P_k|-(k-1)n)$. ProofFirstly consider the $k = 2$ case first: if $P \subset \{1, \cdots, n \}$ and $Q \subset \{1, \cdots, n \}$. Then $|P \cap Q| = |P|+|Q|-|P \cup Q|$ and $|P \cup Q| \geq n$ so a bit of case bashing yields $|P \cap Q| \geq \max(0, |P|+|Q|-n)$. Now induct on $k$, with the base case $k=1, 2$ being done. Then assume it's true till $r$. For the inductive step, let $P = |P_1 \cap P_2 \cap \cdots P_r| \geq \max(0, |P_1|+|P_2|+\cdots+|P_r|-(r-1)n)$ and $Q = P_{r+1}$. Now using this in the $k=2$ case, we see $|P \cap Q| \geq \max(0, |P|+|Q|-n) \geq \max(0, |P_1|+|P_2|+\cdots+|P_r|-(r-1)n+|P_{r+1}|-n) = \max(0, |P_1|+|P_2|+\cdots+|P_r|+|P_{r+1}|-rn)$, so done. Now define $E(\epsilon, b, m)$ for $0 < \epsilon \ll 1$ as the smallest positive integer $N$ such that for all $n \geq N$, we have $\frac{|S(n,b,m)|}{n} < \epsilon$. Such a number always exists by the definition of limits and Lemma-1. Let $M = \max(E(\frac{1}{5000}, 2, m), E(\frac{1}{5000}, 3, m), E(\frac{1}{5000}, 4, m), \cdots, E(\frac{1}{5000}, 1389, m), 100!)$. By definition of $M$, for all $n \geq M$ and for all $2 \leq b \leq 1389$ we have $|T(n,b,m)| \geq \frac{4999n}{5000}$. Define $P_i = T(n, i+1, m)$ for $1 \leq i \leq 1388$. By Lemma-2, $$|P_1 \cap P_2 \cap P_3 \cdots P_{1388}| \geq \max(0, |P_1|+|P_2|+\cdots+|P_{1388}|-1387n) \geq \max(0, \frac{4999*1388n}{5000}-1387n) > .722n$$, so there's atleast one number in the intersection of all these sets. Then this number clearly satisfies the desired property so we're done

Every variable in the solution are positive integers otherwise stated. Let $S(n,b,m) = \{ p | n \geq p \geq 1 \text{ and } s_b(p) \leq m \}$, and $T(n,b,m) = \{ p | n \geq p \geq 1 \text{ and } s_b(p) > m \}$. Note that $|S(n,b,m)|+|T(n,b,m)| = n$ for any $b,m$. Lemma - 1 : For any fixed $b > 1$, $m \geq 0$, we have $\lim_{n \rightarrow \infty} \frac{|S(n,b,m)|}{n} = 0$. ProofLet $f_b(i,j)$ denote the numbers which takes atmost $i$ digits to write in base $b$ and has digit sum exactly equal to $j$. Let $g(i,j)$ denote the number of solutions to the equation $$ x_1 + x_2 + \cdots + x_i = j$$with $x_i \in \mathbb{N}_{0}$. Now note for any $i,j$, we have $f_b(i,j) \leq g(i,j)$ (assume $\max(x_1, \cdots, x_i) \leq b-1$, then interpret $x_1x_2 \cdots x_i$ in base $b$). Also note that $g(i,j) = \binom{i+j-1}{j} = \frac{(i+j-1)(i+j-2)\cdots(i-1)}{j!} < (i+j-1)^j$. Now, WLOG assuming $n > b^{{m+100000}^{m+10000}}$ and letting $i = \lfloor \log_b n \rfloor + 2$ $$ 0 \leq \frac{|S(n,b,m)|}{n} \leq \frac{g(i,0) + g(i,1) + \cdots + g(i, b-1)}{n} < \frac{\sum_{0 \leq j \leq b-1} (i+j-1)^j}{n}$$$$< (i+m-1)^m \frac{m+1}{n} < \frac{(2i)^m}{n} (m+1) \leq \frac{(100 \log_b n)^m (m+1)}{n}$$. Now since $\lim_{n \rightarrow \infty}\frac{(\log_b n)^m}{n} = 0$, by squeeze theorem and noting that $100^m(m+1)$ are constant, we see $\lim_{n \rightarrow \infty} \frac{|S(n,b,m)|}{n} = 0$. Lemma - 2 Let $P_1, P_2, \cdots, P_k \subset \{1, \cdots, n \}$. Then $|P_1 \cap P_2 \cap P_3 \cdots P_k| \geq \max(0, |P_1|+|P_2|+\cdots+|P_k|-(k-1)n)$. ProofFirstly consider the $k = 2$ case first: if $P \subset \{1, \cdots, n \}$ and $Q \subset \{1, \cdots, n \}$. Then $|P \cap Q| = |P|+|Q|-|P \cup Q|$ and $|P \cup Q| \geq n$ so a bit of case bashing yields $|P \cap Q| \geq \max(0, |P|+|Q|-n)$. Now induct on $k$, with the base case $k=1, 2$ being done. Then assume it's true till $r$. For the inductive step, let $P = |P_1 \cap P_2 \cap \cdots P_r| \geq \max(0, |P_1|+|P_2|+\cdots+|P_r|-(r-1)n)$ and $Q = P_{r+1}$. Now using this in the $k=2$ case, we see $|P \cap Q| \geq \max(0, |P|+|Q|-n) \geq \max(0, |P_1|+|P_2|+\cdots+|P_r|-(r-1)n+|P_{r+1}|-n) = \max(0, |P_1|+|P_2|+\cdots+|P_r|+|P_{r+1}|-rn)$, so done. Now define $E(\epsilon, b, m)$ for $0 < \epsilon \ll 1$ as the smallest positive integer $N$ such that for all $n \geq N$, we have $\frac{|S(n,b,m)|}{n} < \epsilon$. Such a number always exists by the definition of limits and Lemma-1. Let $M = \max(E(\frac{1}{5000}, 2, m), E(\frac{1}{5000}, 3, m), E(\frac{1}{5000}, 4, m), \cdots, E(\frac{1}{5000}, 1389, m), 100!)$. By definition of $M$, for all $n \geq M$ and for all $2 \leq b \leq 1389$ we have $|T(n,b,m)| \geq \frac{4999n}{5000}$. Define $P_i = T(n, i+1, m)$ for $1 \leq i \leq 1388$. By Lemma-2, $$|P_1 \cap P_2 \cap P_3 \cdots P_{1388}| \geq \max(0, |P_1|+|P_2|+\cdots+|P_{1388}|-1387n) \geq \max(0, \frac{4999*1388n}{5000}-1387n) > .722n$$, so there's atleast one number in the intersection of all these sets. Then this number clearly satisfies the desired property so we're done