We will show that there exists no positive integer $n$ for which the statement of the problem is satisfied. Let us denote by $A_n$ the set of all rational numbers $r$ for which there exist an integer $b$ and nonzero integers $a_1, \dots, a_n$ such that $r = b + \sum_{k=1}^{n} \frac{1}{a_k}$. In order to establish our claim, it suffices to show that for any positive integer $n$ the set $A_n$ does not exhaust the set of all rational numbers. We will show more: in fact there exists a pair $p_n, q_n$ of rational numbers with $p_n < q_n$ such that there exists no member of the set $A_n$ which lies strictly in between $p_n$ and $q_n$. We will show the existence of such a pair recursively.

For $n = 1$ it suffices to take $p_1 = \frac{1}{3}$ and $q_1 = \frac{1}{2}$. Now suppose, we were able to choose for some $n$ a pair $p_n, q_n$ such that $p_n < q_n$ for which there is no element from the set $A_n$ lying strictly between $p_n$ and $q_n$. Set $d = q_n - p_n$, and define $p_{n+1} = p_n + \frac{d}{3}$ and $q'_{n+1} = q_n - \frac{d}{3}$. Suppose that there exist an integer $b$ and nonzero integers $a_1, \dots, a_{n+1}$ for which
$$p_{n+1} < b + \sum_{k=1}^{n+1} \frac{1}{a_k} < q'_{n+1}$$
holds.

By the induction hypothesis, we have for each $l \in \{1, \dots, n+1\}$, either
$$p_n \ge b + \sum_{k=1, k \neq l}^{n+1} \frac{1}{a_k}$$
or
$$b + \sum_{k=1, k \neq l}^{n+1} \frac{1}{a_k} \ge q_n.$$

If the former holds, then we have $\frac{1}{a_l} < p_{n+1} - p_n = \frac{d}{3}$, while if the latter holds, then $\frac{1}{a_l} < q'_{n+1} - q_n = -\frac{d}{3}$. In either case, we get $-\frac{3}{d} < a_l < \frac{3}{d}$, which implies that for each $l \in \{1, \dots, n+1\}$, there are only finitely many possible integral values that $a_l$ can take. Furthermore if we fix $(a_1, \dots, a_{n+1})$, the number of integers $b$ that satisfy the property $p_{n+1} < b + \sum_{k=1}^{n+1} \frac{1}{a_k} < q'_{n+1}$ is also finite.

Consequently, we can conclude that there exist at most a finite number of elements of the set $A_{n+1}$ which lie in between $p_{n+1}$ and $q'_{n+1}$. If we define $q_{n+1}$ to be equal to the minimum of such exceptional numbers of $A_{n+1}$ if such exists and equal to $q'_{n+1}$ if there is no such exceptional element, then we can see that $p_{n+1}$ and $q_{n+1}$ will satisfy the statement of our assertion for $n+1$, and therefore, our induction process is complete and our assertion is proved.