I guess the OP means that $a_1\leq\dots\leq a_m$. In that case the statement is even true for $s=mn-n+2$.

To see this, define the numbers $1\leq c(i,k)\leq m$ such that $v_i=(a_{c(i,1)},\dots,a_{c(i,n)})$. In addition, consider the integers $w_i:=c(i,1)+\dots+c(i,n)$. Then, for any pair $1\leq i<j\leq s$, we have $v_i\leq v_j$, hence $c(i,k)\leq c(j,k)$ for any $1\leq k\leq n$, hence also $w_i\leq w_j$. Moreover, $w_i=w_j$ can only hold when $c(i,k)=c(j,k)$ for each $1\leq k\leq n$, i.e. when $v_i=v_j$. Now observe that $n\leq w_1\leq\dots\leq w_s\leq mn$, hence for $s=mn-n+2$ we have $w_i=w_j$ for some $i<j$, so $v_i=v_j$ by the discussion above. The proof is complete.

I guess the OP means that the values are ordered linearly, i.e. $a_1<\dots<a_m$. In that case the statement is even true for $s=mn-n+2$.

To see this, define the numbers $1\leq c(i,k)\leq m$ such that $v_i=(a_{c(i,1)},\dots,a_{c(i,n)})$. In addition, consider the integers $w_i:=c(i,1)+\dots+c(i,n)$. Then, for any pair $1\leq i<j\leq s$, we have $v_i\leq v_j$, hence $c(i,k)\leq c(j,k)$ for any $1\leq k\leq n$, hence also $w_i\leq w_j$. Moreover, $w_i=w_j$ can only hold when $c(i,k)=c(j,k)$ for each $1\leq k\leq n$, i.e. when $v_i=v_j$. Now observe that $n\leq w_1\leq\dots\leq w_s\leq mn$, hence for $s=mn-n+2$ we have $w_i=w_j$ for some $i<j$, so $v_i=v_j$ by the discussion above. The proof is complete.

I think the statement is even true for $s=mn-n+2$. Let $w_i$ be the sum of entries of $v_i$. Then clearly $n\leq w_1\leq\dots\leq w_s\leq mn$, hence for $s=mn-n+2$ we have $w_i=w_j$ for some $i<j$. However $v_i\leq v_j$, so we must have $v_i=v_j$ (otherwise we would have $w_i<w_j$).

I guess the OP means that $a_1\leq\dots\leq a_m$. In that case the statement is even true for $s=mn-n+2$.

To see this, define the numbers $1\leq c(i,k)\leq m$ such that $v_i=(a_{c(i,1)},\dots,a_{c(i,n)})$. In addition, consider the integers $w_i:=c(i,1)+\dots+c(i,n)$. Then, for any pair $1\leq i<j\leq s$, we have $v_i\leq v_j$, hence $c(i,k)\leq c(j,k)$ for any $1\leq k\leq n$, hence also $w_i\leq w_j$. Moreover, $w_i=w_j$ can only hold when $c(i,k)=c(j,k)$ for each $1\leq k\leq n$, i.e. when $v_i=v_j$. Now observe that $n\leq w_1\leq\dots\leq w_s\leq mn$, hence for $s=mn-n+2$ we have $w_i=w_j$ for some $i<j$, so $v_i=v_j$ by the discussion above. The proof is complete.