Let $a_1, a_2, a_3, \dots$ be an infinite sequence of positive integers satisfying the conditions of the problem. From $a_{n+1} \le a_n + 5$ ($n = 1, 2, \dots$) it follows that for every $n$ $a_n \le 5 (n-1) + a_1$ is satisfied. In particular, for all sufficiently large $n$ $a_n < 6n$ is satisfied (in fact, $n \ge a_1 - 4$ will do.) As $a_n$ is a multiple of $n$ by assumption, we see that $a_n \le 5n$ for all sufficiently large $n$.

Suppose there exists an $n$ for which $a_n > 5n$. Since there are only a finite number of such $n$'s, as we saw above, there must be the largest such $n$, which we denote by $N$. Then, $a_N > 5N$ and $a_{N+1} \le 5(N + 1)$. As $a_N$ is a multiple of $N$, we in fact have $a_N \ge 6N$. From
$$5 \ge a_N - a_{N+1} \ge 6N - 5(N+1) = N - 5$$
we obtain that $N \le 10$. We have thus shown that if $n \ge 11$, then $a_n \le 5n$. In particular, $a_{11} \le 55$. and since $a_n$ is a multiple of $n$ and does not exceed $a_{n+1} + 5$, we obtain recursively,
$$\begin{align*} a_{10} &\le 60, \quad a_9 \le 63, \quad a_8 \le 64, \quad a_7 \le 63, \quad a_6 \le 66, \\ a_5 &\le 70, \quad a_4 \le 72, \quad a_3 \le 75, \quad a_2 \le 80, \quad a_1 \le 85. \end{align*}$$
If, on the other hand, we define
$$\begin{align*} a_1 &= 85, \quad a_2 = 80, \quad a_3 = 75, \quad a_4 = 72, \quad a_5 = 70, \quad a_6 = 66, \\ a_7 &= 63, \quad a_8 = 64, \quad a_9 = 63, \quad a_{10} = 60, \quad a_n = 5n\ (n \ge 11) \end{align*}$$
then this sequence can be shown to satisfy the conditions of the problem, and hence we conclude that 85 is the desired maximum.