Let $c$ be a nonnegative integer. Find all infinite sequences of positive integers $\{a_{n}\}_{n\geq 1}$ satisfying the following condition: for all positive integer $n$, there are exactly $a_n$ positive integers $i$ such that $a_i\leq a_{n+1}+c$.
1. **Claim:** \(a_i\) is weakly increasing.
   - **Proof:** Suppose \(a_i < a_j\) for some \(i, j\). Then, if \(a_{j+1} \leq a_{i+1}\), we have \(a_{j+1} + c \leq a_{i+1} + c\). This implies that there are at least as many \(k\) with \(a_k \leq a_{i+1} + c\) as there are with \(a_k \leq a_{j+1} + c\), which contradicts the given condition. Therefore, \(a_i > a_{i+1}\) for any \(i\) is impossible, as the sequence would decrease forever, which is not possible since each \(a_i\) is positive. Hence, \(a_i\) is increasing (not necessarily strictly).

2. If \(a_i = a_j\), then \(a_{i-1} = a_{j-1}\) as well. Hence, \(a_1 = a_2 = \dots = a_M\) for some \(M \geq 1\), and then \(a_i\) strictly increases. It cannot be constant throughout because then we run into contradictions for large \(n\).

3. **Claim:** The sequence is (weakly) convex, i.e., \(a_{i+2} - a_{i+1} \geq a_{i+1} - a_i\) for all \(i \geq 1\).
   - **Proof:** Consider some \(n \geq M\). The number of \(i\) such that \(a_{n+1} + c < a_i \leq a_{n+2} + c\) is \(a_{n+1} - a_n\). Since these \(a_i\) are distinct (each of these appear after the first \(M\) values), it follows that \(a_{n+2} - a_{n+1} \geq a_{n+1} - a_n\).

4. If \(a_{i+1} - a_i\) eventually becomes unbounded, then \(a_{k+1} - a_k > c\) for large enough \(k\). As a result, the number of \(i\) such that \(a_i \leq a_{n+1} + c\) becomes \(n + 1\) for large enough \(n\), implying \(a_n = n + 1\) for large enough \(n\). So \(a_{n+1} - a_n = 1\) for large enough \(n\), which is absurd. Therefore, \(a_{n+1} - a_n\) eventually converges to some fixed positive integer, say \(b\).

5. Suppose \(a_{n+1} = a_n + b\) for all \(n \geq N\). Then the number of \(i\) with \(a_i \leq a_{n+1} + c\) for some fixed \(n \geq N\) becomes:
   \[
   a_n = n + 1 + \left\lfloor \frac{c}{b} \right\rfloor
   \]
   which implies that \(b = a_{N+1} - a_N = 1\).

6. Putting all the previous observations together, we obtain that \(a_1 = a_2 = \dots = a_M = d\) (some constant) and then \(a_i\) increases by 1 every term. That is, \(a_i = d\) for \(i \leq M\) and \(a_i = d + i - M\) for \(i > M\). From before, we have \(a_n = n + 1 + c\) for large enough \(n\), and so \(d - M = c + 1\).

7. **Claim:** It is not possible to have \(M > 1\), that is, we must have \(M = 1\).
   - **Proof:** If \(M > 1\), then there are exactly \(d\) indices \(i\) with \(a_i \leq a_M + c = a_d + c\). But, there are exactly \(M + c\) such indices (all \(i \in [1, M + c]\)), hence \(d = M + c\), contradicting \(d = M + c + 1\). So we must have \(M = 1\).

8. Since \(M = 1\), \(d = c + 2\) and we have \(a_i = i + (c + 1)\) for all \(i \geq 1\). It is easy to see that this sequence indeed satisfies the problem's constraints.

The final answer is \(\boxed{a_n = n + (c + 1)}\).