Shadows of partitions of lcm

Question

$\DeclareMathOperator\lcm{lcm}$Fix an integer $n\geq1$. Denote the least common multiple $L_n=\lcm(1,2,\dots,n)$.

QUESTION. Is the following true? For each integer partition $\lambda=(\lambda_1,\lambda_2,\dots)\vdash 2L_n$ whose parts satisfy $\lambda_i\leq n$ there exists a partition $\mu=(\mu_1,\mu_2,\dots)\vdash L_n$ such that for each part $\mu_j$ there is a part $\lambda_i$ for which $\mu_j=\lambda_i$ for some $i$. The latter condition means $$\{\mu_1,\mu_2,\dots\} \subseteq \{\lambda_1,\lambda_2,\dots\}$$ when regarded as multisets.

Remark. Another way to look at this is: the problem requires splitting a partition of $2L_n$ into two partitions of $L_n$.

Postscript. Just a remark: I'm convinced that instead of Ilya's construction "at most $(k-1)$ of coin $k$", we could work also with "at most $(k-2)$ of coin $k$". Two points on the latter choice: (1) the argument works without the constraint for "large values of $n$"; (2) it makes any presence of both coins $k=1$ and $k=2$ to be essential.

Answer 1

We shall explain our algorithm. Assume we have coins of values among $1,2,\dots, n$ with some multiplicities, and for a total value of $2L_n$. That is, $$\sum_{i=1}^n i\cdot b_i=2L_n \qquad \text{for some $b_i\geq0$}.$$ Our aim is to pick some of them into the pocket so that the total value in the pocket will be $L_n$. Again, that means, there exist number $0\leq a_i\leq b_i$ such that $$\sum_{i=1}^n i\cdot a_i=L_n.$$ Say that a coin $k$ is essential if there are at least $k-1$ coins of value $k$ outside the pocket; otherwise, we say $k$ is inessential. Notice that this notion of essential is dynamic, a coin may become inessential during the process!

At any stage, we maintain that the total $T$ in our pocket will be divisible by the largest essential number. At the beginning, this holds by vacua.

Now, assume that at some moment (perhaps, at the very start) the two largest essential numbers are $k>\ell$. By our condition, $T$ is divisible by $k$; then we can take some (nonzero number of) coins of value $k$ so that the new $T$ will be divisible by $\mathop{\mathrm{lcm}}(k,\ell)$ (for simplicity, say we take those coins one by one), We have sufficiently many such coins to reach that, as $k$ is essential. After that, the largest essential number becomes either $k$ or $\ell$, so in any case we preserve the desired property. Then we repeat the step, using the new two largest essential numbers.

If $k$ is the only remaining essential number, we merely take coin of value $k$ at that step.

Notice that at any moment when we take coin of value $a$ the total in the pocket is divisible by $a$. This means that we cannot jump over $L_n$: if, at the end, the total in the pocket is at least $L_n$, then at some moment we had exactly $L_n$, as desired.

Suppose we run out of essential coins with fewer than a total of $L_n$ in our pocket. Surely, the process stops. But, we claim this event can not occur. Indeed, since all coins outside pocket are inessential, the total value of the coins outside the pocket does not exceed \begin{align*} n(n-2)&+(n-1)(n-3)+\dots+2\cdot 0 =\frac{(n-1)(n-2)(2n+3)}6 \\ &\leq \frac{n(n-1)(n-2)}2\leq\mathop{\mathrm{lcm}}(n,n-1,n-2)\leq L_n \end{align*} for $n\geq1$. The last equality is due to this result. This leads to a contradiction because both set of coins (inside and outside our pocket) are not larger than $L_n$, with one inequality being strict. Hence the total is less than $2L_n$. The proof is now complete.