Solution:

The sum of the angles of an $n$-gon is $(n-2) 180$, so the average angle measure is $(n-2) 180 / n$. The common difference in this arithmetic progression is at least $1$, so the difference between the largest and smallest angles is at least $n-1$. So the largest angle is at least $(n-1)/2 + (n-2) 180 / n$. Since the polygon is convex, this quantity is no larger than $179$: $(n-1)/2 - 360/n \leq -1$, so that $360/n - n/2 \geq 1/2$. Multiplying by $2n$ gives $720 - n^{2} \geq n$. So $n(n+1) \leq 720$, which forces $n \leq 26$.

Of course, since the common difference is an integer, and the angle measures are integers, $(n-2) 180 / n$ must be an integer or a half integer, so $(n-2) 360 / n = 360 - 720 / n$ is an integer, and then $720 / n$ must be an integer. This leaves only $n = 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24$ as possibilities.

When $n$ is even, $(n-2) 180 / n$ is not an angle of the polygon, but the mean of the two middle angles. So the common difference is at least $2$ when $(n-2) 180 / n$ is an integer. For $n=20$, the middle angle is $162$, so the largest angle is at least $162 + 38/2 = 181$, since $38$ is no larger than the difference between the smallest and largest angles. For $n=24$, the middle angle is $165$, again leading to a contradiction. So no solution exists for $n=20,24$. All of the others possess solutions: