Assume that $a_{n+1} \ge a_{n+2}$ holds for some positive integers $n$. Since any positive integer $i$ with $a_i \le a_{n+2} + c$ must also satisfy $a_i \le a_{n+1} + c$, it follows that $a_n \ge a_{n+1}$. Hence, if $a_n < a_{n+1}$ holds for some positive integer $n$, then $a_{n+1} < a_{n+2}$ follows, and by induction, we have $a_n < a_{n+1} < a_{n+2} < a_{n+3} < \dots$.

Similarly, when $a_n > a_{n+1}$ holds for some positive integer $n$, we have $a_n > a_{n+1} > a_{n+2} > \dots$, especially $a_{n+d} \le a_n - d$ for any non-negative integer $d$. This leads to a contradiction since $a_{n+a_n} \le 0$.

Suppose that $a_n = a_{n+1}$ for all positive integers $n$. Then, as all integers $i \ge 2$ satisfy $a_i = a_2 \le a_2 + c$, this contradicts the fact that there are exactly $a_1$ positive integers $i$ satisfying it. Therefore, there exists a positive integer $k$ such that $a_k < a_{k+1}$, and from the above discussion, we have $a_1 \le a_2 \le \dots \le a_k < a_{k+1} < a_{k+2} < \dots$.

For integers $n \ge k$, since any integer $i > n+c+1$ satisfies $a_i > a_{n+c+1} \ge a_{n+1} + c$, we have $a_n \le n+c+1$. Therefore, if we set $b_n = a_n - n$ ($n \ge k$), then we have $b_n \le c+1$ and $b_k \le b_{k+1} \le b_{k+2} \le \dots$. Thus, there exist an integer $d$ and an integer $M \ge k$ such that for $n \ge M$, $b_n = d$, i.e., $a_n = n+d$. Since $a_1 \le a_2 \le \dots \le a_{M+c+1} < a_{M+c+2} < \dots$, for any positive integer $i$, $a_i \le a_{M+1}+c = a_{M+c+1}$ and $i \le M+c+1$ are equivalent. Therefore, $a_M = M+c+1$, which is also equal to $M+d$, and for integers $n \ge M$, we have $a_n = n+c+1$.

Suppose that for an integer $N \ge 2$, we have $a_n = n+c+1$ for all $n \ge N$. Since $a_1 \le a_2 \le \dots \le a_{N+c} < a_{N+c+1} < \dots$, for any positive integer $i$, $a_i \le a_N + c = a_{N+c}$ and $i \le N+c$ are equivalent. Therefore, $a_{N-1} = N+c$.

Thus, by induction, we have $a_n = n+c+1$ for any positive integer $n$. This indeed satisfies the problem condition because $a_i \le a_{n+1} + c$ and $i \le a_n$ are equivalent for any positive integer $i$.