What is the acceleration of a particle at its highest point when thrown vertically upwards?

a downward acceleration

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I think you are subconsciously mixing up velocity with acceleration. Let me give you an example. Imagine these are the measured speeds of a particle thrown vertically into the air at different times:

time, speed 0s, 50m/s 1s, 40m/s 2s, 30m/s 3s, 20m/s 4s, 10m/s 5s, 0m/s 6s, -10m/s 7s, -20m/s 8s, -30m/s 9s, -40m/s 10s, -50m/s <-- it hits the ground

Here a minus sign simply means the particle is coming towards the earth. Now the acceleration at for example 4s is the difference between the speeds at 3s and at 5s divided by the elapsed time, which is (20m/s-0m/s)/2s=10m/s^2

Similarly, the acceleration at 5s is the difference between the speeds at 4s and at 6s divided by the elapsed time, which is (10m/s-(-10m/s))/2s=10m/s^s

This motion is called a motion with constant acceleration, as the earth exerts a constant force on the particle at all times, no matter where it is (unless the particle gets really really far, in which case the force won't be constant any more).